How to antidifferentiate with a negative exponent?












2












$begingroup$


$6x^{-2} + 2x^{-4} -3x^{-3}$.



How would the anti diff process work for fractions? I have $frac{1}{6x^2}$ for the first part. Am I supposed to add one to x, and then put it all over 3? How would the fraction then resolve?






calculus







share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    $6x^{-2} + 2x^{-4} -3x^{-3}$.



    How would the anti diff process work for fractions? I have $frac{1}{6x^2}$ for the first part. Am I supposed to add one to x, and then put it all over 3? How would the fraction then resolve?






    calculus







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      $6x^{-2} + 2x^{-4} -3x^{-3}$.



      How would the anti diff process work for fractions? I have $frac{1}{6x^2}$ for the first part. Am I supposed to add one to x, and then put it all over 3? How would the fraction then resolve?






      calculus







      share|cite|improve this question









      $endgroup$




      $6x^{-2} + 2x^{-4} -3x^{-3}$.



      How would the anti diff process work for fractions? I have $frac{1}{6x^2}$ for the first part. Am I supposed to add one to x, and then put it all over 3? How would the fraction then resolve?






      calculus




      calculus






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked May 9 '16 at 13:51









      user6050977user6050977

      1658




      1658






















          4 Answers
          4






          active

          oldest

          votes


















          4












          $begingroup$

          Hint. A general rule, working for all exponents (both negative and non-negative):




          $$
          f(x)=x^{alpha} quad text{gives an antiderivative } , F(x)=frac{x^{alpha+1}}{alpha+1}+C quad text{if} quad alpha neq-1,
          $$
          $$
          f(x)=x^{-1}= frac1xquad text{gives an antiderivative } , F(x)=ln (x)+C quad text{if} quad x>0,
          $$




          where $C$ is any constant.






          share|cite|improve this answer











          $endgroup$





















            3












            $begingroup$

            Two general formulas:



            $(x^n)' = n cdot x^{n-1}$ (for any integer $n neq 0$)

            $(c cdot f(x))' = c cdot f'(x)$ (for any number $c$)



            So we know that: $(x^{-1})' = -1 cdot x^{-2}$

            So: $(-6 cdot x^{-1})' = -6 cdot (-1) cdot x^{-2} = 6 cdot x^{-2}$

            So the antiderivative of $6 cdot x^{-2}$ is $-6 cdot x^{-1}$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Isn't subtracting from an exponent supposed to be for differentiating? And adding one to an exponent for antidifferentiating?
              $endgroup$
              – user6050977
              May 9 '16 at 17:24






            • 1




              $begingroup$
              @user6050977 That's correct.
              $endgroup$
              – peter.petrov
              May 9 '16 at 20:15










            • $begingroup$
              I don.t understand. So, it's $6 / x^2$? Then add one to x for $x^{-1}$, then dividing by that same -1, for $-6x^{-1}$?
              $endgroup$
              – user6050977
              May 10 '16 at 9:50










            • $begingroup$
              Yes. I don't understand what you don't understand. Just sit down and write it slowly and think about it, using the formulas on this page.
              $endgroup$
              – peter.petrov
              May 10 '16 at 10:42



















            1












            $begingroup$

            Just continue to use the rule $(x^{n})'=nx^{n-1}$ for all $nin mathbb{Z}$, $nnot=0$. Hence,



            $(x^{-1})'=-x^{-2}implies(-6x^{-1})'=6x^{-2}$.



            $(x^{-3})'=-3x^{-4}implies(-frac{2}{3}x^{-3})'=2x^{-4}$.



            $(x^{-2})'=-2x^{-3}implies(frac{3}{2}x^{-2})'=-3x^{-3}$.



            So, the antiderivative is: $-6x^{-1}-frac{2}{3}x^{-3}+frac{3}{2}x^{-2}$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Isn't subtracting from an exponent supposed to be for differentiating? And adding one to an exponent for antidifferentiating?
              $endgroup$
              – user6050977
              May 9 '16 at 17:24






            • 1




              $begingroup$
              Yes. The exponents you have are $-2, -4, -3$. When you add one to each of them, you get $-1, -3, -2$.
              $endgroup$
              – ervx
              May 9 '16 at 19:11










            • $begingroup$
              I don.t understand. So, it's $6 / x^2$? Then add one to x for $x^{-1}$, then dividing by that same -1, for $-6x^{-1}$?
              $endgroup$
              – user6050977
              May 10 '16 at 9:49












            • $begingroup$
              You have $6/x^{2}=6x^{-2}$. Add $1$ to the exponent and consider $x^{-1}$. We have that $(x^{-1})'=-x^{-2}$, but we need something whose derivative is $6x^{-2}$, so, since $(x^{-1})'=-x^{-2}$, it must be that $(-6x^{-1})'=6x^{-2}$.
              $endgroup$
              – ervx
              May 10 '16 at 13:34



















            0












            $begingroup$

            I think your confusion comes from missing this:



            $frac{1}{x^2}=x^{-2}$
            $frac{1}{x^3}=x^{-3}$
            $frac{1}{x^4}=x^{-4}$



            Just proceed as you would for positive powers, but being careful about minus signs.



            $frac{1}{x}=x^{-1} $ is a special case since you can't get it by differentiating $x^0$ (try it!) but they've very kindly avoided putting that in the expression.



            Example:



            Find the antiderivative of $frac{2}{x^5}$.



            First write this as $$frac{2}{x^5}=2x^{-5}$$



            Thinking of this as $2x^{n}$, what is $n$? $$n=-5$$



            So, what is $n+1$? $$n+1=-4$$



            In which case, what is $frac{x^{n+1}}{n+1}$?



            $$frac{x^{n+1}}{n+1}=frac{x^{-4}}{-4} = -frac{1}{4{x^4}}$$



            But we wanted the antiderivative of $2x^{-5}$, not just $x^{-5}$, so we still need to multiply by 2 and we finally get $$2left(-frac{1}{4{x^4}}right)=-frac{1}{2{x^4}}$$






            share|cite|improve this answer











            $endgroup$













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              4 Answers
              4






              active

              oldest

              votes








              4 Answers
              4






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

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              4












              $begingroup$

              Hint. A general rule, working for all exponents (both negative and non-negative):




              $$
              f(x)=x^{alpha} quad text{gives an antiderivative } , F(x)=frac{x^{alpha+1}}{alpha+1}+C quad text{if} quad alpha neq-1,
              $$
              $$
              f(x)=x^{-1}= frac1xquad text{gives an antiderivative } , F(x)=ln (x)+C quad text{if} quad x>0,
              $$




              where $C$ is any constant.






              share|cite|improve this answer











              $endgroup$


















                4












                $begingroup$

                Hint. A general rule, working for all exponents (both negative and non-negative):




                $$
                f(x)=x^{alpha} quad text{gives an antiderivative } , F(x)=frac{x^{alpha+1}}{alpha+1}+C quad text{if} quad alpha neq-1,
                $$
                $$
                f(x)=x^{-1}= frac1xquad text{gives an antiderivative } , F(x)=ln (x)+C quad text{if} quad x>0,
                $$




                where $C$ is any constant.






                share|cite|improve this answer











                $endgroup$
















                  4












                  4








                  4





                  $begingroup$

                  Hint. A general rule, working for all exponents (both negative and non-negative):




                  $$
                  f(x)=x^{alpha} quad text{gives an antiderivative } , F(x)=frac{x^{alpha+1}}{alpha+1}+C quad text{if} quad alpha neq-1,
                  $$
                  $$
                  f(x)=x^{-1}= frac1xquad text{gives an antiderivative } , F(x)=ln (x)+C quad text{if} quad x>0,
                  $$




                  where $C$ is any constant.






                  share|cite|improve this answer











                  $endgroup$



                  Hint. A general rule, working for all exponents (both negative and non-negative):




                  $$
                  f(x)=x^{alpha} quad text{gives an antiderivative } , F(x)=frac{x^{alpha+1}}{alpha+1}+C quad text{if} quad alpha neq-1,
                  $$
                  $$
                  f(x)=x^{-1}= frac1xquad text{gives an antiderivative } , F(x)=ln (x)+C quad text{if} quad x>0,
                  $$




                  where $C$ is any constant.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Nov 29 '18 at 21:35









                  Brahadeesh

                  6,19742361




                  6,19742361










                  answered May 9 '16 at 14:01









                  Olivier OloaOlivier Oloa

                  108k17176293




                  108k17176293























                      3












                      $begingroup$

                      Two general formulas:



                      $(x^n)' = n cdot x^{n-1}$ (for any integer $n neq 0$)

                      $(c cdot f(x))' = c cdot f'(x)$ (for any number $c$)



                      So we know that: $(x^{-1})' = -1 cdot x^{-2}$

                      So: $(-6 cdot x^{-1})' = -6 cdot (-1) cdot x^{-2} = 6 cdot x^{-2}$

                      So the antiderivative of $6 cdot x^{-2}$ is $-6 cdot x^{-1}$.






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        Isn't subtracting from an exponent supposed to be for differentiating? And adding one to an exponent for antidifferentiating?
                        $endgroup$
                        – user6050977
                        May 9 '16 at 17:24






                      • 1




                        $begingroup$
                        @user6050977 That's correct.
                        $endgroup$
                        – peter.petrov
                        May 9 '16 at 20:15










                      • $begingroup$
                        I don.t understand. So, it's $6 / x^2$? Then add one to x for $x^{-1}$, then dividing by that same -1, for $-6x^{-1}$?
                        $endgroup$
                        – user6050977
                        May 10 '16 at 9:50










                      • $begingroup$
                        Yes. I don't understand what you don't understand. Just sit down and write it slowly and think about it, using the formulas on this page.
                        $endgroup$
                        – peter.petrov
                        May 10 '16 at 10:42
















                      3












                      $begingroup$

                      Two general formulas:



                      $(x^n)' = n cdot x^{n-1}$ (for any integer $n neq 0$)

                      $(c cdot f(x))' = c cdot f'(x)$ (for any number $c$)



                      So we know that: $(x^{-1})' = -1 cdot x^{-2}$

                      So: $(-6 cdot x^{-1})' = -6 cdot (-1) cdot x^{-2} = 6 cdot x^{-2}$

                      So the antiderivative of $6 cdot x^{-2}$ is $-6 cdot x^{-1}$.






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        Isn't subtracting from an exponent supposed to be for differentiating? And adding one to an exponent for antidifferentiating?
                        $endgroup$
                        – user6050977
                        May 9 '16 at 17:24






                      • 1




                        $begingroup$
                        @user6050977 That's correct.
                        $endgroup$
                        – peter.petrov
                        May 9 '16 at 20:15










                      • $begingroup$
                        I don.t understand. So, it's $6 / x^2$? Then add one to x for $x^{-1}$, then dividing by that same -1, for $-6x^{-1}$?
                        $endgroup$
                        – user6050977
                        May 10 '16 at 9:50










                      • $begingroup$
                        Yes. I don't understand what you don't understand. Just sit down and write it slowly and think about it, using the formulas on this page.
                        $endgroup$
                        – peter.petrov
                        May 10 '16 at 10:42














                      3












                      3








                      3





                      $begingroup$

                      Two general formulas:



                      $(x^n)' = n cdot x^{n-1}$ (for any integer $n neq 0$)

                      $(c cdot f(x))' = c cdot f'(x)$ (for any number $c$)



                      So we know that: $(x^{-1})' = -1 cdot x^{-2}$

                      So: $(-6 cdot x^{-1})' = -6 cdot (-1) cdot x^{-2} = 6 cdot x^{-2}$

                      So the antiderivative of $6 cdot x^{-2}$ is $-6 cdot x^{-1}$.






                      share|cite|improve this answer











                      $endgroup$



                      Two general formulas:



                      $(x^n)' = n cdot x^{n-1}$ (for any integer $n neq 0$)

                      $(c cdot f(x))' = c cdot f'(x)$ (for any number $c$)



                      So we know that: $(x^{-1})' = -1 cdot x^{-2}$

                      So: $(-6 cdot x^{-1})' = -6 cdot (-1) cdot x^{-2} = 6 cdot x^{-2}$

                      So the antiderivative of $6 cdot x^{-2}$ is $-6 cdot x^{-1}$.







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited May 9 '16 at 14:01

























                      answered May 9 '16 at 13:54









                      peter.petrovpeter.petrov

                      5,417821




                      5,417821












                      • $begingroup$
                        Isn't subtracting from an exponent supposed to be for differentiating? And adding one to an exponent for antidifferentiating?
                        $endgroup$
                        – user6050977
                        May 9 '16 at 17:24






                      • 1




                        $begingroup$
                        @user6050977 That's correct.
                        $endgroup$
                        – peter.petrov
                        May 9 '16 at 20:15










                      • $begingroup$
                        I don.t understand. So, it's $6 / x^2$? Then add one to x for $x^{-1}$, then dividing by that same -1, for $-6x^{-1}$?
                        $endgroup$
                        – user6050977
                        May 10 '16 at 9:50










                      • $begingroup$
                        Yes. I don't understand what you don't understand. Just sit down and write it slowly and think about it, using the formulas on this page.
                        $endgroup$
                        – peter.petrov
                        May 10 '16 at 10:42


















                      • $begingroup$
                        Isn't subtracting from an exponent supposed to be for differentiating? And adding one to an exponent for antidifferentiating?
                        $endgroup$
                        – user6050977
                        May 9 '16 at 17:24






                      • 1




                        $begingroup$
                        @user6050977 That's correct.
                        $endgroup$
                        – peter.petrov
                        May 9 '16 at 20:15










                      • $begingroup$
                        I don.t understand. So, it's $6 / x^2$? Then add one to x for $x^{-1}$, then dividing by that same -1, for $-6x^{-1}$?
                        $endgroup$
                        – user6050977
                        May 10 '16 at 9:50










                      • $begingroup$
                        Yes. I don't understand what you don't understand. Just sit down and write it slowly and think about it, using the formulas on this page.
                        $endgroup$
                        – peter.petrov
                        May 10 '16 at 10:42
















                      $begingroup$
                      Isn't subtracting from an exponent supposed to be for differentiating? And adding one to an exponent for antidifferentiating?
                      $endgroup$
                      – user6050977
                      May 9 '16 at 17:24




                      $begingroup$
                      Isn't subtracting from an exponent supposed to be for differentiating? And adding one to an exponent for antidifferentiating?
                      $endgroup$
                      – user6050977
                      May 9 '16 at 17:24




                      1




                      1




                      $begingroup$
                      @user6050977 That's correct.
                      $endgroup$
                      – peter.petrov
                      May 9 '16 at 20:15




                      $begingroup$
                      @user6050977 That's correct.
                      $endgroup$
                      – peter.petrov
                      May 9 '16 at 20:15












                      $begingroup$
                      I don.t understand. So, it's $6 / x^2$? Then add one to x for $x^{-1}$, then dividing by that same -1, for $-6x^{-1}$?
                      $endgroup$
                      – user6050977
                      May 10 '16 at 9:50




                      $begingroup$
                      I don.t understand. So, it's $6 / x^2$? Then add one to x for $x^{-1}$, then dividing by that same -1, for $-6x^{-1}$?
                      $endgroup$
                      – user6050977
                      May 10 '16 at 9:50












                      $begingroup$
                      Yes. I don't understand what you don't understand. Just sit down and write it slowly and think about it, using the formulas on this page.
                      $endgroup$
                      – peter.petrov
                      May 10 '16 at 10:42




                      $begingroup$
                      Yes. I don't understand what you don't understand. Just sit down and write it slowly and think about it, using the formulas on this page.
                      $endgroup$
                      – peter.petrov
                      May 10 '16 at 10:42











                      1












                      $begingroup$

                      Just continue to use the rule $(x^{n})'=nx^{n-1}$ for all $nin mathbb{Z}$, $nnot=0$. Hence,



                      $(x^{-1})'=-x^{-2}implies(-6x^{-1})'=6x^{-2}$.



                      $(x^{-3})'=-3x^{-4}implies(-frac{2}{3}x^{-3})'=2x^{-4}$.



                      $(x^{-2})'=-2x^{-3}implies(frac{3}{2}x^{-2})'=-3x^{-3}$.



                      So, the antiderivative is: $-6x^{-1}-frac{2}{3}x^{-3}+frac{3}{2}x^{-2}$.






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        Isn't subtracting from an exponent supposed to be for differentiating? And adding one to an exponent for antidifferentiating?
                        $endgroup$
                        – user6050977
                        May 9 '16 at 17:24






                      • 1




                        $begingroup$
                        Yes. The exponents you have are $-2, -4, -3$. When you add one to each of them, you get $-1, -3, -2$.
                        $endgroup$
                        – ervx
                        May 9 '16 at 19:11










                      • $begingroup$
                        I don.t understand. So, it's $6 / x^2$? Then add one to x for $x^{-1}$, then dividing by that same -1, for $-6x^{-1}$?
                        $endgroup$
                        – user6050977
                        May 10 '16 at 9:49












                      • $begingroup$
                        You have $6/x^{2}=6x^{-2}$. Add $1$ to the exponent and consider $x^{-1}$. We have that $(x^{-1})'=-x^{-2}$, but we need something whose derivative is $6x^{-2}$, so, since $(x^{-1})'=-x^{-2}$, it must be that $(-6x^{-1})'=6x^{-2}$.
                        $endgroup$
                        – ervx
                        May 10 '16 at 13:34
















                      1












                      $begingroup$

                      Just continue to use the rule $(x^{n})'=nx^{n-1}$ for all $nin mathbb{Z}$, $nnot=0$. Hence,



                      $(x^{-1})'=-x^{-2}implies(-6x^{-1})'=6x^{-2}$.



                      $(x^{-3})'=-3x^{-4}implies(-frac{2}{3}x^{-3})'=2x^{-4}$.



                      $(x^{-2})'=-2x^{-3}implies(frac{3}{2}x^{-2})'=-3x^{-3}$.



                      So, the antiderivative is: $-6x^{-1}-frac{2}{3}x^{-3}+frac{3}{2}x^{-2}$.






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        Isn't subtracting from an exponent supposed to be for differentiating? And adding one to an exponent for antidifferentiating?
                        $endgroup$
                        – user6050977
                        May 9 '16 at 17:24






                      • 1




                        $begingroup$
                        Yes. The exponents you have are $-2, -4, -3$. When you add one to each of them, you get $-1, -3, -2$.
                        $endgroup$
                        – ervx
                        May 9 '16 at 19:11










                      • $begingroup$
                        I don.t understand. So, it's $6 / x^2$? Then add one to x for $x^{-1}$, then dividing by that same -1, for $-6x^{-1}$?
                        $endgroup$
                        – user6050977
                        May 10 '16 at 9:49












                      • $begingroup$
                        You have $6/x^{2}=6x^{-2}$. Add $1$ to the exponent and consider $x^{-1}$. We have that $(x^{-1})'=-x^{-2}$, but we need something whose derivative is $6x^{-2}$, so, since $(x^{-1})'=-x^{-2}$, it must be that $(-6x^{-1})'=6x^{-2}$.
                        $endgroup$
                        – ervx
                        May 10 '16 at 13:34














                      1












                      1








                      1





                      $begingroup$

                      Just continue to use the rule $(x^{n})'=nx^{n-1}$ for all $nin mathbb{Z}$, $nnot=0$. Hence,



                      $(x^{-1})'=-x^{-2}implies(-6x^{-1})'=6x^{-2}$.



                      $(x^{-3})'=-3x^{-4}implies(-frac{2}{3}x^{-3})'=2x^{-4}$.



                      $(x^{-2})'=-2x^{-3}implies(frac{3}{2}x^{-2})'=-3x^{-3}$.



                      So, the antiderivative is: $-6x^{-1}-frac{2}{3}x^{-3}+frac{3}{2}x^{-2}$.






                      share|cite|improve this answer









                      $endgroup$



                      Just continue to use the rule $(x^{n})'=nx^{n-1}$ for all $nin mathbb{Z}$, $nnot=0$. Hence,



                      $(x^{-1})'=-x^{-2}implies(-6x^{-1})'=6x^{-2}$.



                      $(x^{-3})'=-3x^{-4}implies(-frac{2}{3}x^{-3})'=2x^{-4}$.



                      $(x^{-2})'=-2x^{-3}implies(frac{3}{2}x^{-2})'=-3x^{-3}$.



                      So, the antiderivative is: $-6x^{-1}-frac{2}{3}x^{-3}+frac{3}{2}x^{-2}$.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered May 9 '16 at 14:01









                      ervxervx

                      10.3k31338




                      10.3k31338












                      • $begingroup$
                        Isn't subtracting from an exponent supposed to be for differentiating? And adding one to an exponent for antidifferentiating?
                        $endgroup$
                        – user6050977
                        May 9 '16 at 17:24






                      • 1




                        $begingroup$
                        Yes. The exponents you have are $-2, -4, -3$. When you add one to each of them, you get $-1, -3, -2$.
                        $endgroup$
                        – ervx
                        May 9 '16 at 19:11










                      • $begingroup$
                        I don.t understand. So, it's $6 / x^2$? Then add one to x for $x^{-1}$, then dividing by that same -1, for $-6x^{-1}$?
                        $endgroup$
                        – user6050977
                        May 10 '16 at 9:49












                      • $begingroup$
                        You have $6/x^{2}=6x^{-2}$. Add $1$ to the exponent and consider $x^{-1}$. We have that $(x^{-1})'=-x^{-2}$, but we need something whose derivative is $6x^{-2}$, so, since $(x^{-1})'=-x^{-2}$, it must be that $(-6x^{-1})'=6x^{-2}$.
                        $endgroup$
                        – ervx
                        May 10 '16 at 13:34


















                      • $begingroup$
                        Isn't subtracting from an exponent supposed to be for differentiating? And adding one to an exponent for antidifferentiating?
                        $endgroup$
                        – user6050977
                        May 9 '16 at 17:24






                      • 1




                        $begingroup$
                        Yes. The exponents you have are $-2, -4, -3$. When you add one to each of them, you get $-1, -3, -2$.
                        $endgroup$
                        – ervx
                        May 9 '16 at 19:11










                      • $begingroup$
                        I don.t understand. So, it's $6 / x^2$? Then add one to x for $x^{-1}$, then dividing by that same -1, for $-6x^{-1}$?
                        $endgroup$
                        – user6050977
                        May 10 '16 at 9:49












                      • $begingroup$
                        You have $6/x^{2}=6x^{-2}$. Add $1$ to the exponent and consider $x^{-1}$. We have that $(x^{-1})'=-x^{-2}$, but we need something whose derivative is $6x^{-2}$, so, since $(x^{-1})'=-x^{-2}$, it must be that $(-6x^{-1})'=6x^{-2}$.
                        $endgroup$
                        – ervx
                        May 10 '16 at 13:34
















                      $begingroup$
                      Isn't subtracting from an exponent supposed to be for differentiating? And adding one to an exponent for antidifferentiating?
                      $endgroup$
                      – user6050977
                      May 9 '16 at 17:24




                      $begingroup$
                      Isn't subtracting from an exponent supposed to be for differentiating? And adding one to an exponent for antidifferentiating?
                      $endgroup$
                      – user6050977
                      May 9 '16 at 17:24




                      1




                      1




                      $begingroup$
                      Yes. The exponents you have are $-2, -4, -3$. When you add one to each of them, you get $-1, -3, -2$.
                      $endgroup$
                      – ervx
                      May 9 '16 at 19:11




                      $begingroup$
                      Yes. The exponents you have are $-2, -4, -3$. When you add one to each of them, you get $-1, -3, -2$.
                      $endgroup$
                      – ervx
                      May 9 '16 at 19:11












                      $begingroup$
                      I don.t understand. So, it's $6 / x^2$? Then add one to x for $x^{-1}$, then dividing by that same -1, for $-6x^{-1}$?
                      $endgroup$
                      – user6050977
                      May 10 '16 at 9:49






                      $begingroup$
                      I don.t understand. So, it's $6 / x^2$? Then add one to x for $x^{-1}$, then dividing by that same -1, for $-6x^{-1}$?
                      $endgroup$
                      – user6050977
                      May 10 '16 at 9:49














                      $begingroup$
                      You have $6/x^{2}=6x^{-2}$. Add $1$ to the exponent and consider $x^{-1}$. We have that $(x^{-1})'=-x^{-2}$, but we need something whose derivative is $6x^{-2}$, so, since $(x^{-1})'=-x^{-2}$, it must be that $(-6x^{-1})'=6x^{-2}$.
                      $endgroup$
                      – ervx
                      May 10 '16 at 13:34




                      $begingroup$
                      You have $6/x^{2}=6x^{-2}$. Add $1$ to the exponent and consider $x^{-1}$. We have that $(x^{-1})'=-x^{-2}$, but we need something whose derivative is $6x^{-2}$, so, since $(x^{-1})'=-x^{-2}$, it must be that $(-6x^{-1})'=6x^{-2}$.
                      $endgroup$
                      – ervx
                      May 10 '16 at 13:34











                      0












                      $begingroup$

                      I think your confusion comes from missing this:



                      $frac{1}{x^2}=x^{-2}$
                      $frac{1}{x^3}=x^{-3}$
                      $frac{1}{x^4}=x^{-4}$



                      Just proceed as you would for positive powers, but being careful about minus signs.



                      $frac{1}{x}=x^{-1} $ is a special case since you can't get it by differentiating $x^0$ (try it!) but they've very kindly avoided putting that in the expression.



                      Example:



                      Find the antiderivative of $frac{2}{x^5}$.



                      First write this as $$frac{2}{x^5}=2x^{-5}$$



                      Thinking of this as $2x^{n}$, what is $n$? $$n=-5$$



                      So, what is $n+1$? $$n+1=-4$$



                      In which case, what is $frac{x^{n+1}}{n+1}$?



                      $$frac{x^{n+1}}{n+1}=frac{x^{-4}}{-4} = -frac{1}{4{x^4}}$$



                      But we wanted the antiderivative of $2x^{-5}$, not just $x^{-5}$, so we still need to multiply by 2 and we finally get $$2left(-frac{1}{4{x^4}}right)=-frac{1}{2{x^4}}$$






                      share|cite|improve this answer











                      $endgroup$


















                        0












                        $begingroup$

                        I think your confusion comes from missing this:



                        $frac{1}{x^2}=x^{-2}$
                        $frac{1}{x^3}=x^{-3}$
                        $frac{1}{x^4}=x^{-4}$



                        Just proceed as you would for positive powers, but being careful about minus signs.



                        $frac{1}{x}=x^{-1} $ is a special case since you can't get it by differentiating $x^0$ (try it!) but they've very kindly avoided putting that in the expression.



                        Example:



                        Find the antiderivative of $frac{2}{x^5}$.



                        First write this as $$frac{2}{x^5}=2x^{-5}$$



                        Thinking of this as $2x^{n}$, what is $n$? $$n=-5$$



                        So, what is $n+1$? $$n+1=-4$$



                        In which case, what is $frac{x^{n+1}}{n+1}$?



                        $$frac{x^{n+1}}{n+1}=frac{x^{-4}}{-4} = -frac{1}{4{x^4}}$$



                        But we wanted the antiderivative of $2x^{-5}$, not just $x^{-5}$, so we still need to multiply by 2 and we finally get $$2left(-frac{1}{4{x^4}}right)=-frac{1}{2{x^4}}$$






                        share|cite|improve this answer











                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          I think your confusion comes from missing this:



                          $frac{1}{x^2}=x^{-2}$
                          $frac{1}{x^3}=x^{-3}$
                          $frac{1}{x^4}=x^{-4}$



                          Just proceed as you would for positive powers, but being careful about minus signs.



                          $frac{1}{x}=x^{-1} $ is a special case since you can't get it by differentiating $x^0$ (try it!) but they've very kindly avoided putting that in the expression.



                          Example:



                          Find the antiderivative of $frac{2}{x^5}$.



                          First write this as $$frac{2}{x^5}=2x^{-5}$$



                          Thinking of this as $2x^{n}$, what is $n$? $$n=-5$$



                          So, what is $n+1$? $$n+1=-4$$



                          In which case, what is $frac{x^{n+1}}{n+1}$?



                          $$frac{x^{n+1}}{n+1}=frac{x^{-4}}{-4} = -frac{1}{4{x^4}}$$



                          But we wanted the antiderivative of $2x^{-5}$, not just $x^{-5}$, so we still need to multiply by 2 and we finally get $$2left(-frac{1}{4{x^4}}right)=-frac{1}{2{x^4}}$$






                          share|cite|improve this answer











                          $endgroup$



                          I think your confusion comes from missing this:



                          $frac{1}{x^2}=x^{-2}$
                          $frac{1}{x^3}=x^{-3}$
                          $frac{1}{x^4}=x^{-4}$



                          Just proceed as you would for positive powers, but being careful about minus signs.



                          $frac{1}{x}=x^{-1} $ is a special case since you can't get it by differentiating $x^0$ (try it!) but they've very kindly avoided putting that in the expression.



                          Example:



                          Find the antiderivative of $frac{2}{x^5}$.



                          First write this as $$frac{2}{x^5}=2x^{-5}$$



                          Thinking of this as $2x^{n}$, what is $n$? $$n=-5$$



                          So, what is $n+1$? $$n+1=-4$$



                          In which case, what is $frac{x^{n+1}}{n+1}$?



                          $$frac{x^{n+1}}{n+1}=frac{x^{-4}}{-4} = -frac{1}{4{x^4}}$$



                          But we wanted the antiderivative of $2x^{-5}$, not just $x^{-5}$, so we still need to multiply by 2 and we finally get $$2left(-frac{1}{4{x^4}}right)=-frac{1}{2{x^4}}$$







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Nov 30 '18 at 0:07

























                          answered Nov 29 '18 at 22:19









                          timtfjtimtfj

                          1,318318




                          1,318318






























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