Show that $f(x) = sqrt{x^3}+3$ for ${x≥0 | x in mathbb{R}}$ is differentiable and find its derivative.












1












$begingroup$


I'd like to find out if the function $f(x) = sqrt{x^3}+3$ is differentiable for ${x≥0 | x in mathbb{R}}$ and if it is, then find it's derivative.



First I "simplified" the function $f(x)= (sqrt {x^3}+3)^frac{1}{2}$ and used the difference quotient to show that $f(x)$ is differentiable for $x>0$.



$$frac{f(x) - f(x_0)}{x-x_0} = frac{(sqrt{x^3}+3)- left(sqrt{x_0^3}+3 right)}{x-x_0}= frac{sqrt{x^3}- sqrt{x_0^3}}{x-x_0}$$



We know that $x^3-x_0^3=left(sqrt{x^3} - sqrt{x_0^3}right)left(sqrt{x^3} + sqrt{x_0^3}right)$, therefore:



$$ implies frac{left(sqrt{x^3} - sqrt{x_0^3} right)}{left({x^3} - sqrt{x_0^3}right)left({x^3} + sqrt{x_0^3}right)} = frac{1}{left({x^3} + sqrt{x_0^3}right)}$$



Additionally, we show that $lim_limits{xto x_0}left({x^3} + sqrt{x_0^3}right) = 2sqrt {x_0^3}$ and therefore:



$$implies f'(x) = frac{1}{2sqrt {x_0^3}} $$



Therefore when ${x>0 | x in mathbb{R}}$, $f(x)$ is differentiable and now I'd have to show that $f(x)$ is differentiable for $x=0$. I do this by using the difference quotient.



$$lim _limits{x to 0}frac{f(x) - f(0)}{x-0}=lim _limits{x to 0}frac{sqrt{x^3}+3 }{x}=frac{lim_limits{x to 0} sqrt {x^3}+3}{lim_limits{x to 0}x}$$



Since I get a situation where I would have to divide by $0$ I can't continue but in my exercise sheet it says that its $∞$ and I don't understand why. From the looks of it, I made a mistake somewhere or I have some understanding gap.






real-analysis calculus limits derivatives







share|cite|improve this question











$endgroup$












  • $begingroup$
    Oh snap. That seems to be a writing mistake. Sorry about that. Yeah now that I look back at that, I'm not sure how I can even use that fact to help with the proof.
    $endgroup$
    – Ski Mask
    Dec 13 '18 at 18:20










  • $begingroup$
    Chain rule maybe? Then treat $x=0$ apart
    $endgroup$
    – Federico
    Dec 13 '18 at 18:21


















1












$begingroup$


I'd like to find out if the function $f(x) = sqrt{x^3}+3$ is differentiable for ${x≥0 | x in mathbb{R}}$ and if it is, then find it's derivative.



First I "simplified" the function $f(x)= (sqrt {x^3}+3)^frac{1}{2}$ and used the difference quotient to show that $f(x)$ is differentiable for $x>0$.



$$frac{f(x) - f(x_0)}{x-x_0} = frac{(sqrt{x^3}+3)- left(sqrt{x_0^3}+3 right)}{x-x_0}= frac{sqrt{x^3}- sqrt{x_0^3}}{x-x_0}$$



We know that $x^3-x_0^3=left(sqrt{x^3} - sqrt{x_0^3}right)left(sqrt{x^3} + sqrt{x_0^3}right)$, therefore:



$$ implies frac{left(sqrt{x^3} - sqrt{x_0^3} right)}{left({x^3} - sqrt{x_0^3}right)left({x^3} + sqrt{x_0^3}right)} = frac{1}{left({x^3} + sqrt{x_0^3}right)}$$



Additionally, we show that $lim_limits{xto x_0}left({x^3} + sqrt{x_0^3}right) = 2sqrt {x_0^3}$ and therefore:



$$implies f'(x) = frac{1}{2sqrt {x_0^3}} $$



Therefore when ${x>0 | x in mathbb{R}}$, $f(x)$ is differentiable and now I'd have to show that $f(x)$ is differentiable for $x=0$. I do this by using the difference quotient.



$$lim _limits{x to 0}frac{f(x) - f(0)}{x-0}=lim _limits{x to 0}frac{sqrt{x^3}+3 }{x}=frac{lim_limits{x to 0} sqrt {x^3}+3}{lim_limits{x to 0}x}$$



Since I get a situation where I would have to divide by $0$ I can't continue but in my exercise sheet it says that its $∞$ and I don't understand why. From the looks of it, I made a mistake somewhere or I have some understanding gap.






real-analysis calculus limits derivatives







share|cite|improve this question











$endgroup$












  • $begingroup$
    Oh snap. That seems to be a writing mistake. Sorry about that. Yeah now that I look back at that, I'm not sure how I can even use that fact to help with the proof.
    $endgroup$
    – Ski Mask
    Dec 13 '18 at 18:20










  • $begingroup$
    Chain rule maybe? Then treat $x=0$ apart
    $endgroup$
    – Federico
    Dec 13 '18 at 18:21
















1












1








1





$begingroup$


I'd like to find out if the function $f(x) = sqrt{x^3}+3$ is differentiable for ${x≥0 | x in mathbb{R}}$ and if it is, then find it's derivative.



First I "simplified" the function $f(x)= (sqrt {x^3}+3)^frac{1}{2}$ and used the difference quotient to show that $f(x)$ is differentiable for $x>0$.



$$frac{f(x) - f(x_0)}{x-x_0} = frac{(sqrt{x^3}+3)- left(sqrt{x_0^3}+3 right)}{x-x_0}= frac{sqrt{x^3}- sqrt{x_0^3}}{x-x_0}$$



We know that $x^3-x_0^3=left(sqrt{x^3} - sqrt{x_0^3}right)left(sqrt{x^3} + sqrt{x_0^3}right)$, therefore:



$$ implies frac{left(sqrt{x^3} - sqrt{x_0^3} right)}{left({x^3} - sqrt{x_0^3}right)left({x^3} + sqrt{x_0^3}right)} = frac{1}{left({x^3} + sqrt{x_0^3}right)}$$



Additionally, we show that $lim_limits{xto x_0}left({x^3} + sqrt{x_0^3}right) = 2sqrt {x_0^3}$ and therefore:



$$implies f'(x) = frac{1}{2sqrt {x_0^3}} $$



Therefore when ${x>0 | x in mathbb{R}}$, $f(x)$ is differentiable and now I'd have to show that $f(x)$ is differentiable for $x=0$. I do this by using the difference quotient.



$$lim _limits{x to 0}frac{f(x) - f(0)}{x-0}=lim _limits{x to 0}frac{sqrt{x^3}+3 }{x}=frac{lim_limits{x to 0} sqrt {x^3}+3}{lim_limits{x to 0}x}$$



Since I get a situation where I would have to divide by $0$ I can't continue but in my exercise sheet it says that its $∞$ and I don't understand why. From the looks of it, I made a mistake somewhere or I have some understanding gap.






real-analysis calculus limits derivatives







share|cite|improve this question











$endgroup$




I'd like to find out if the function $f(x) = sqrt{x^3}+3$ is differentiable for ${x≥0 | x in mathbb{R}}$ and if it is, then find it's derivative.



First I "simplified" the function $f(x)= (sqrt {x^3}+3)^frac{1}{2}$ and used the difference quotient to show that $f(x)$ is differentiable for $x>0$.



$$frac{f(x) - f(x_0)}{x-x_0} = frac{(sqrt{x^3}+3)- left(sqrt{x_0^3}+3 right)}{x-x_0}= frac{sqrt{x^3}- sqrt{x_0^3}}{x-x_0}$$



We know that $x^3-x_0^3=left(sqrt{x^3} - sqrt{x_0^3}right)left(sqrt{x^3} + sqrt{x_0^3}right)$, therefore:



$$ implies frac{left(sqrt{x^3} - sqrt{x_0^3} right)}{left({x^3} - sqrt{x_0^3}right)left({x^3} + sqrt{x_0^3}right)} = frac{1}{left({x^3} + sqrt{x_0^3}right)}$$



Additionally, we show that $lim_limits{xto x_0}left({x^3} + sqrt{x_0^3}right) = 2sqrt {x_0^3}$ and therefore:



$$implies f'(x) = frac{1}{2sqrt {x_0^3}} $$



Therefore when ${x>0 | x in mathbb{R}}$, $f(x)$ is differentiable and now I'd have to show that $f(x)$ is differentiable for $x=0$. I do this by using the difference quotient.



$$lim _limits{x to 0}frac{f(x) - f(0)}{x-0}=lim _limits{x to 0}frac{sqrt{x^3}+3 }{x}=frac{lim_limits{x to 0} sqrt {x^3}+3}{lim_limits{x to 0}x}$$



Since I get a situation where I would have to divide by $0$ I can't continue but in my exercise sheet it says that its $∞$ and I don't understand why. From the looks of it, I made a mistake somewhere or I have some understanding gap.






real-analysis calculus limits derivatives




real-analysis calculus limits derivatives






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 13 '18 at 18:18









Ethan Bolker

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44k552117










asked Dec 13 '18 at 18:04









Ski MaskSki Mask

661318




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  • $begingroup$
    Oh snap. That seems to be a writing mistake. Sorry about that. Yeah now that I look back at that, I'm not sure how I can even use that fact to help with the proof.
    $endgroup$
    – Ski Mask
    Dec 13 '18 at 18:20










  • $begingroup$
    Chain rule maybe? Then treat $x=0$ apart
    $endgroup$
    – Federico
    Dec 13 '18 at 18:21




















  • $begingroup$
    Oh snap. That seems to be a writing mistake. Sorry about that. Yeah now that I look back at that, I'm not sure how I can even use that fact to help with the proof.
    $endgroup$
    – Ski Mask
    Dec 13 '18 at 18:20










  • $begingroup$
    Chain rule maybe? Then treat $x=0$ apart
    $endgroup$
    – Federico
    Dec 13 '18 at 18:21


















$begingroup$
Oh snap. That seems to be a writing mistake. Sorry about that. Yeah now that I look back at that, I'm not sure how I can even use that fact to help with the proof.
$endgroup$
– Ski Mask
Dec 13 '18 at 18:20




$begingroup$
Oh snap. That seems to be a writing mistake. Sorry about that. Yeah now that I look back at that, I'm not sure how I can even use that fact to help with the proof.
$endgroup$
– Ski Mask
Dec 13 '18 at 18:20












$begingroup$
Chain rule maybe? Then treat $x=0$ apart
$endgroup$
– Federico
Dec 13 '18 at 18:21






$begingroup$
Chain rule maybe? Then treat $x=0$ apart
$endgroup$
– Federico
Dec 13 '18 at 18:21












1 Answer
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$begingroup$

You start off fine, somewhere your algebra goes a little wonky.



You are not quite explict with that algebra. So, I am having a had time putting my finger on the error.



This is how I would do it.



$f'(a) = lim_limits{xto a} frac {f(x) - f(a)}{x-a}\
lim_limits{xto a} frac {(sqrt{x^3}+3) - (sqrt{a^3}+3)}{x-a}\
lim_limits{xto a} frac {sqrt{x^3} - sqrt{a^3}}{x-a}\
lim_limits{xto a} frac {(sqrt{x^3} - sqrt{a^3})(sqrt{x^3} + sqrt{a^3})}{(x-a)(sqrt{x^3} + sqrt{a^3})}\
lim_limits{xto a} frac {x^3 - a^3}{(x-a)(sqrt{x^3} + sqrt{a^3})}\
lim_limits{xto a} frac {(x-a)(x^2+ax+a^2)}{(x-a)(sqrt{x^3} + sqrt{a^3})}\
f'(a) =frac {3a^2}{2asqrt{a}}\
f'(a) =frac {3sqrt a}{2}\
f'(0) =0\
$






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    $begingroup$

    You start off fine, somewhere your algebra goes a little wonky.



    You are not quite explict with that algebra. So, I am having a had time putting my finger on the error.



    This is how I would do it.



    $f'(a) = lim_limits{xto a} frac {f(x) - f(a)}{x-a}\
    lim_limits{xto a} frac {(sqrt{x^3}+3) - (sqrt{a^3}+3)}{x-a}\
    lim_limits{xto a} frac {sqrt{x^3} - sqrt{a^3}}{x-a}\
    lim_limits{xto a} frac {(sqrt{x^3} - sqrt{a^3})(sqrt{x^3} + sqrt{a^3})}{(x-a)(sqrt{x^3} + sqrt{a^3})}\
    lim_limits{xto a} frac {x^3 - a^3}{(x-a)(sqrt{x^3} + sqrt{a^3})}\
    lim_limits{xto a} frac {(x-a)(x^2+ax+a^2)}{(x-a)(sqrt{x^3} + sqrt{a^3})}\
    f'(a) =frac {3a^2}{2asqrt{a}}\
    f'(a) =frac {3sqrt a}{2}\
    f'(0) =0\
    $






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      You start off fine, somewhere your algebra goes a little wonky.



      You are not quite explict with that algebra. So, I am having a had time putting my finger on the error.



      This is how I would do it.



      $f'(a) = lim_limits{xto a} frac {f(x) - f(a)}{x-a}\
      lim_limits{xto a} frac {(sqrt{x^3}+3) - (sqrt{a^3}+3)}{x-a}\
      lim_limits{xto a} frac {sqrt{x^3} - sqrt{a^3}}{x-a}\
      lim_limits{xto a} frac {(sqrt{x^3} - sqrt{a^3})(sqrt{x^3} + sqrt{a^3})}{(x-a)(sqrt{x^3} + sqrt{a^3})}\
      lim_limits{xto a} frac {x^3 - a^3}{(x-a)(sqrt{x^3} + sqrt{a^3})}\
      lim_limits{xto a} frac {(x-a)(x^2+ax+a^2)}{(x-a)(sqrt{x^3} + sqrt{a^3})}\
      f'(a) =frac {3a^2}{2asqrt{a}}\
      f'(a) =frac {3sqrt a}{2}\
      f'(0) =0\
      $






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        You start off fine, somewhere your algebra goes a little wonky.



        You are not quite explict with that algebra. So, I am having a had time putting my finger on the error.



        This is how I would do it.



        $f'(a) = lim_limits{xto a} frac {f(x) - f(a)}{x-a}\
        lim_limits{xto a} frac {(sqrt{x^3}+3) - (sqrt{a^3}+3)}{x-a}\
        lim_limits{xto a} frac {sqrt{x^3} - sqrt{a^3}}{x-a}\
        lim_limits{xto a} frac {(sqrt{x^3} - sqrt{a^3})(sqrt{x^3} + sqrt{a^3})}{(x-a)(sqrt{x^3} + sqrt{a^3})}\
        lim_limits{xto a} frac {x^3 - a^3}{(x-a)(sqrt{x^3} + sqrt{a^3})}\
        lim_limits{xto a} frac {(x-a)(x^2+ax+a^2)}{(x-a)(sqrt{x^3} + sqrt{a^3})}\
        f'(a) =frac {3a^2}{2asqrt{a}}\
        f'(a) =frac {3sqrt a}{2}\
        f'(0) =0\
        $






        share|cite|improve this answer









        $endgroup$



        You start off fine, somewhere your algebra goes a little wonky.



        You are not quite explict with that algebra. So, I am having a had time putting my finger on the error.



        This is how I would do it.



        $f'(a) = lim_limits{xto a} frac {f(x) - f(a)}{x-a}\
        lim_limits{xto a} frac {(sqrt{x^3}+3) - (sqrt{a^3}+3)}{x-a}\
        lim_limits{xto a} frac {sqrt{x^3} - sqrt{a^3}}{x-a}\
        lim_limits{xto a} frac {(sqrt{x^3} - sqrt{a^3})(sqrt{x^3} + sqrt{a^3})}{(x-a)(sqrt{x^3} + sqrt{a^3})}\
        lim_limits{xto a} frac {x^3 - a^3}{(x-a)(sqrt{x^3} + sqrt{a^3})}\
        lim_limits{xto a} frac {(x-a)(x^2+ax+a^2)}{(x-a)(sqrt{x^3} + sqrt{a^3})}\
        f'(a) =frac {3a^2}{2asqrt{a}}\
        f'(a) =frac {3sqrt a}{2}\
        f'(0) =0\
        $







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 13 '18 at 18:25









        Doug MDoug M

        45.3k31954




        45.3k31954






























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