# A different implicit differentiation

• Oct 19th 2008, 09:51 PM
Mike5055
A different implicit differentiation
Alright, so I'm having some problems. I am supposed to find dy/dx in terms of x and y by implicit differentiation. I wasn't doing too bad until I hit two problems:

Problem 1) y^2 + ye^x + e^2x=3
Problem 2) e^y^2 - x^2 - y^2=0

For the first one, I am not certain how to handle the ones to the power of x and 2x when the first one is y to the power of two. I might be overthinking it though.

For the second one, I am not certain how to handle the e^y^2, so I am just confused with that one.

Any help I can get would be much appreciated!
• Oct 19th 2008, 09:54 PM
kalagota
Quote:

Originally Posted by Mike5055
Alright, so I'm having some problems. I am supposed to find dy/dx in terms of x and y by implicit differentiation. I wasn't doing too bad until I hit two problems:

Problem 1) y^2 + ye^x + e^2x=3
Problem 2) e^y^2 - x^2 - y^2=0

For the first one, I am not certain how to handle the ones to the power of x and 2x when the first one is y to the power of two. I might be overthinking it though.

For the second one, I am not certain how to handle the e^y^2, so I am just confused with that one.

Any help I can get would be much appreciated!

so how about showing us your solution and we'll check for you..
• Oct 19th 2008, 09:58 PM
Mike5055
Alright. I couldn't get to the solution, but for the first one, what I thought is turn it into

2y + 2e(x') + 2xe^x = 3

and for the second one, if you derive it to

(y^2)(e) - 2x - 2y = 0

I'm not certain if that was the correct step to make with them or not.
• Oct 19th 2008, 10:07 PM
kalagota
ok, here are some hints..

$D_x[y] = \frac{dy}{dx}$ since $y$ is a variable too.. thus, $D_x[y^2] = 2y\cdot D_x[y] = 2y\cdot \frac{dy}{dx}$ by the power and chain rules..

now, for $ye^x$, you may want to use product rule..

and, $e^{2x}$, try using chain rule..

Ü
• Oct 19th 2008, 10:09 PM
Mike5055
Thank you! I'll go through and give that a try!
• Oct 19th 2008, 11:21 PM
dax918
1) y^2 + ye^x + e^(2x) = 3

2y(dy/dx) + e^x(dy/dx) + ye^x + 2e^(2x) = 0

2y(dy/dx) + e^x(dy/dx) = -ye^x - 2e^(2x)

(dy/dx) = [-ye^x - 2e^(2x)] / [ 2y + e^x ]

2) e^(y^2) - x^2 - y^2 = 0

2ye^(y^2)(dy/dx) - 2x - 2y(dy/dx) = 0

(dy/dx)(2ye^(y^2) - 2y) = 2x

(dy/dx) = x / [ ye^(y^2) - y]

Edited: Sorry, I didn't see your last post, I agree, give it a try by yourself first.