# Thread: (chain rule) implicit differentition

1. ## (chain rule) implicit differentition

using chain rule F(x,y)= y^5+x^2 y^3-ye^x^2-1=0
thank you again

2. Originally Posted by fearless901
using chain rule F(x,y)= y^5+x^2 y^3-ye^x^2-1=0
thank you again
Differentiate this function implicitly? but with respect to x or y?

3. Originally Posted by pickslides
Differentiate this function implicitly? but with respect to x or y?
compute using dy/dx using chain rule

4. Originally Posted by fearless901
using chain rule F(x,y)= y^5+x^2 y^3-ye^x^2-1=0
thank you again
$\displaystyle \frac{dy}{dx}=-\frac{F_x}{F_y}=-\frac{2xy^3-2xye^{x^2}}{5y^4+3x^2y^2-e^{x^2}}$

5. ty theemptyset, but can u tell me how you come about the answer?? thank you so much, i stink at math, but i like it

6. nvm i got

7. Originally Posted by fearless901
nvm i got
Really? TheEmptySet actually used a shortcut from multivariable Calculus. He didn't do it by implicit differentiation and then solving for dy/dx. If you would like to, show us how you did the problem so others may use this thread in the future as a reference.

8. well i just did it seperately
using:
dF/dx= 2xy^3-2xye^x^2

dF/dy= 5y^4+3x^2y^2-e^x^2

not sure it helped, basically the same thing as TheEmptySet

9. Originally Posted by fearless901
ty theemptyset, but can u tell me how you come about the answer?? thank you so much, i stink at math, but i like it
Try this link, tis a goodie.

http://archives.math.utk.edu/visual....licit.7/4.html

10. Oh ok. You didn't do implicit differentiation. You used partial differentiation. That is fine of course I just assumed by the thread title that you were in a Calc I or II class, not Calc III.

11. yea im in calc III, sry for the misunderstanding

12. Originally Posted by pickslides
Try this link, tis a goodie.

Visual Calculus - Implicit Differentiation
Nice website thxs, BOOKMARKED

13. Originally Posted by Jameson
Really? TheEmptySet actually used a shortcut from multivariable Calculus. He didn't do it by implicit differentiation and then solving for dy/dx. If you would like to, show us how you did the problem so others may use this thread in the future as a reference.
eeppp I need to pay more attention. I thought that is what we wanted. I will add this for completeness

Suppose $\displaystyle y=f(x)$ and that

$\displaystyle F(x,y)=0$ then we get

$\displaystyle F(x,y)=0$ then by implict differentation and the chain rule we get

$\displaystyle \frac{dF}{dx}\cdot \frac{dx}{dx}+\frac{dF}{dy}\frac{dy}{dx}=0$

$\displaystyle F_x+F_y\frac{dy}{dx}=0 \iff F_y\frac{dy}{dx}=-F_x \iff \frac{dy}{dx}=-\frac{F_x}{F_y}$