Implicit Differentiation with a Tangent Line












2












$begingroup$


I was looking to implicitly differentiate $$-22x^6+4x^{33}y+y^7=-17$$
and found it to be $$dfrac{dy}{dx}=dfrac{132x^5-132x^{32}y}{4x^{33}+7y^6}$$Now, I am trying to find the equation of the tangent line to the curve at the coordinate (1,1). So I then plug both 1 in for x and y into the above equation and come up with $$dfrac{0}{11}$$Now I go to solve $$y-y1=m(x-x1)$$ getting $$y-1=0(x-1)$$
resulting in $y=1$ and the equation to be $y=x+1$ for my final answer. Am I going about this in the correct manner?






calculus implicit-differentiation







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  • $begingroup$
    You start with $-22x^6 + 4x^{33}y + y^7 = -17$. Then you take (implicit) derivatives. What you wrote isn't that, and is not what you mean. What you wrote is that you started with the differential equation $y'-22x^6+4x^{33}y + y^7 = -17$.
    $endgroup$
    – Arturo Magidin
    Dec 11 '18 at 21:04






  • 1




    $begingroup$
    I follow your work to get $y=1$ why do you say $y =x+1$? You have shown that the slope is $0$!
    $endgroup$
    – Doug M
    Dec 11 '18 at 21:20
















2












$begingroup$


I was looking to implicitly differentiate $$-22x^6+4x^{33}y+y^7=-17$$
and found it to be $$dfrac{dy}{dx}=dfrac{132x^5-132x^{32}y}{4x^{33}+7y^6}$$Now, I am trying to find the equation of the tangent line to the curve at the coordinate (1,1). So I then plug both 1 in for x and y into the above equation and come up with $$dfrac{0}{11}$$Now I go to solve $$y-y1=m(x-x1)$$ getting $$y-1=0(x-1)$$
resulting in $y=1$ and the equation to be $y=x+1$ for my final answer. Am I going about this in the correct manner?






calculus implicit-differentiation







share|cite|improve this question











$endgroup$












  • $begingroup$
    You start with $-22x^6 + 4x^{33}y + y^7 = -17$. Then you take (implicit) derivatives. What you wrote isn't that, and is not what you mean. What you wrote is that you started with the differential equation $y'-22x^6+4x^{33}y + y^7 = -17$.
    $endgroup$
    – Arturo Magidin
    Dec 11 '18 at 21:04






  • 1




    $begingroup$
    I follow your work to get $y=1$ why do you say $y =x+1$? You have shown that the slope is $0$!
    $endgroup$
    – Doug M
    Dec 11 '18 at 21:20














2












2








2





$begingroup$


I was looking to implicitly differentiate $$-22x^6+4x^{33}y+y^7=-17$$
and found it to be $$dfrac{dy}{dx}=dfrac{132x^5-132x^{32}y}{4x^{33}+7y^6}$$Now, I am trying to find the equation of the tangent line to the curve at the coordinate (1,1). So I then plug both 1 in for x and y into the above equation and come up with $$dfrac{0}{11}$$Now I go to solve $$y-y1=m(x-x1)$$ getting $$y-1=0(x-1)$$
resulting in $y=1$ and the equation to be $y=x+1$ for my final answer. Am I going about this in the correct manner?






calculus implicit-differentiation







share|cite|improve this question











$endgroup$




I was looking to implicitly differentiate $$-22x^6+4x^{33}y+y^7=-17$$
and found it to be $$dfrac{dy}{dx}=dfrac{132x^5-132x^{32}y}{4x^{33}+7y^6}$$Now, I am trying to find the equation of the tangent line to the curve at the coordinate (1,1). So I then plug both 1 in for x and y into the above equation and come up with $$dfrac{0}{11}$$Now I go to solve $$y-y1=m(x-x1)$$ getting $$y-1=0(x-1)$$
resulting in $y=1$ and the equation to be $y=x+1$ for my final answer. Am I going about this in the correct manner?






calculus implicit-differentiation




calculus implicit-differentiation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 11 '18 at 21:17







pijoborde

















asked Dec 11 '18 at 20:59









pijobordepijoborde

376




376












  • $begingroup$
    You start with $-22x^6 + 4x^{33}y + y^7 = -17$. Then you take (implicit) derivatives. What you wrote isn't that, and is not what you mean. What you wrote is that you started with the differential equation $y'-22x^6+4x^{33}y + y^7 = -17$.
    $endgroup$
    – Arturo Magidin
    Dec 11 '18 at 21:04






  • 1




    $begingroup$
    I follow your work to get $y=1$ why do you say $y =x+1$? You have shown that the slope is $0$!
    $endgroup$
    – Doug M
    Dec 11 '18 at 21:20


















  • $begingroup$
    You start with $-22x^6 + 4x^{33}y + y^7 = -17$. Then you take (implicit) derivatives. What you wrote isn't that, and is not what you mean. What you wrote is that you started with the differential equation $y'-22x^6+4x^{33}y + y^7 = -17$.
    $endgroup$
    – Arturo Magidin
    Dec 11 '18 at 21:04






  • 1




    $begingroup$
    I follow your work to get $y=1$ why do you say $y =x+1$? You have shown that the slope is $0$!
    $endgroup$
    – Doug M
    Dec 11 '18 at 21:20
















$begingroup$
You start with $-22x^6 + 4x^{33}y + y^7 = -17$. Then you take (implicit) derivatives. What you wrote isn't that, and is not what you mean. What you wrote is that you started with the differential equation $y'-22x^6+4x^{33}y + y^7 = -17$.
$endgroup$
– Arturo Magidin
Dec 11 '18 at 21:04




$begingroup$
You start with $-22x^6 + 4x^{33}y + y^7 = -17$. Then you take (implicit) derivatives. What you wrote isn't that, and is not what you mean. What you wrote is that you started with the differential equation $y'-22x^6+4x^{33}y + y^7 = -17$.
$endgroup$
– Arturo Magidin
Dec 11 '18 at 21:04




1




1




$begingroup$
I follow your work to get $y=1$ why do you say $y =x+1$? You have shown that the slope is $0$!
$endgroup$
– Doug M
Dec 11 '18 at 21:20




$begingroup$
I follow your work to get $y=1$ why do you say $y =x+1$? You have shown that the slope is $0$!
$endgroup$
– Doug M
Dec 11 '18 at 21:20










2 Answers
2






active

oldest

votes


















1












$begingroup$

Your work is fine but since $m=left(frac{dy}{dx}right)_{(1,1)}=0$ we have



$$y-y_1=m(x-x_1)=0 implies y=1$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    You're right! I for some reason thought we had to maintain the x for the sake of the equation. Thank you.
    $endgroup$
    – pijoborde
    Dec 11 '18 at 21:26










  • $begingroup$
    @pijoborde You are welcome! Bye
    $endgroup$
    – gimusi
    Dec 11 '18 at 21:27



















1












$begingroup$

your calculations for the slope of tangent are correct. The only point that you have missed is the last step of finding the equation of tangent line which is simply $y=1$ not $y=x+1$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    That's exactly the point!
    $endgroup$
    – gimusi
    Dec 11 '18 at 21:27











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Your work is fine but since $m=left(frac{dy}{dx}right)_{(1,1)}=0$ we have



$$y-y_1=m(x-x_1)=0 implies y=1$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    You're right! I for some reason thought we had to maintain the x for the sake of the equation. Thank you.
    $endgroup$
    – pijoborde
    Dec 11 '18 at 21:26










  • $begingroup$
    @pijoborde You are welcome! Bye
    $endgroup$
    – gimusi
    Dec 11 '18 at 21:27
















1












$begingroup$

Your work is fine but since $m=left(frac{dy}{dx}right)_{(1,1)}=0$ we have



$$y-y_1=m(x-x_1)=0 implies y=1$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    You're right! I for some reason thought we had to maintain the x for the sake of the equation. Thank you.
    $endgroup$
    – pijoborde
    Dec 11 '18 at 21:26










  • $begingroup$
    @pijoborde You are welcome! Bye
    $endgroup$
    – gimusi
    Dec 11 '18 at 21:27














1












1








1





$begingroup$

Your work is fine but since $m=left(frac{dy}{dx}right)_{(1,1)}=0$ we have



$$y-y_1=m(x-x_1)=0 implies y=1$$






share|cite|improve this answer









$endgroup$



Your work is fine but since $m=left(frac{dy}{dx}right)_{(1,1)}=0$ we have



$$y-y_1=m(x-x_1)=0 implies y=1$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 11 '18 at 21:22









gimusigimusi

92.9k84494




92.9k84494












  • $begingroup$
    You're right! I for some reason thought we had to maintain the x for the sake of the equation. Thank you.
    $endgroup$
    – pijoborde
    Dec 11 '18 at 21:26










  • $begingroup$
    @pijoborde You are welcome! Bye
    $endgroup$
    – gimusi
    Dec 11 '18 at 21:27


















  • $begingroup$
    You're right! I for some reason thought we had to maintain the x for the sake of the equation. Thank you.
    $endgroup$
    – pijoborde
    Dec 11 '18 at 21:26










  • $begingroup$
    @pijoborde You are welcome! Bye
    $endgroup$
    – gimusi
    Dec 11 '18 at 21:27
















$begingroup$
You're right! I for some reason thought we had to maintain the x for the sake of the equation. Thank you.
$endgroup$
– pijoborde
Dec 11 '18 at 21:26




$begingroup$
You're right! I for some reason thought we had to maintain the x for the sake of the equation. Thank you.
$endgroup$
– pijoborde
Dec 11 '18 at 21:26












$begingroup$
@pijoborde You are welcome! Bye
$endgroup$
– gimusi
Dec 11 '18 at 21:27




$begingroup$
@pijoborde You are welcome! Bye
$endgroup$
– gimusi
Dec 11 '18 at 21:27











1












$begingroup$

your calculations for the slope of tangent are correct. The only point that you have missed is the last step of finding the equation of tangent line which is simply $y=1$ not $y=x+1$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    That's exactly the point!
    $endgroup$
    – gimusi
    Dec 11 '18 at 21:27
















1












$begingroup$

your calculations for the slope of tangent are correct. The only point that you have missed is the last step of finding the equation of tangent line which is simply $y=1$ not $y=x+1$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    That's exactly the point!
    $endgroup$
    – gimusi
    Dec 11 '18 at 21:27














1












1








1





$begingroup$

your calculations for the slope of tangent are correct. The only point that you have missed is the last step of finding the equation of tangent line which is simply $y=1$ not $y=x+1$






share|cite|improve this answer











$endgroup$



your calculations for the slope of tangent are correct. The only point that you have missed is the last step of finding the equation of tangent line which is simply $y=1$ not $y=x+1$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 11 '18 at 22:21

























answered Dec 11 '18 at 21:24









Mohammad Riazi-KermaniMohammad Riazi-Kermani

41.6k42061




41.6k42061












  • $begingroup$
    That's exactly the point!
    $endgroup$
    – gimusi
    Dec 11 '18 at 21:27


















  • $begingroup$
    That's exactly the point!
    $endgroup$
    – gimusi
    Dec 11 '18 at 21:27
















$begingroup$
That's exactly the point!
$endgroup$
– gimusi
Dec 11 '18 at 21:27




$begingroup$
That's exactly the point!
$endgroup$
– gimusi
Dec 11 '18 at 21:27


















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