Implicit Differentiation: Finding dy/dx for cos(y^2) = x^4

[mlowery]

Find $$\frac{dy}{dx}$$ given $$cos(y^2) = x^4$$
Is this correct:

1. $$cos(y^2) = x^4$$

2. $$-sin(y^2) \times 2y \frac{dy}{dx} = 4x^3$$

3. $$\frac{dy}{dx} = \frac{4x^3}{-2sin(y^2)}$$

 

[Doc Al]

Oops... Looks like you left out a "y" in the last step. Write it like this:
$$\frac{dy}{dx} = -\frac{2x^3}{y \sin(y^2)}$$

 

[mlowery]

Yes, that is what I originally did. The thing is, this is a multiple choice question. Of the choices, the only answers close to this are: (Please don't think I am using this forum for answers. I just believe none of the choices are correct).

A) $$\frac{4x^3}{-sin(y^2)}$$

B) $$\frac{4x^3}{-2ysin(y^2)}$$

Here are the other choices:

C) $$\frac{\sqrt{xy}-y}{2xy}$$

D) $$\frac{x^4}{-sin(y^2)}$$

E) $$\frac{4x^3}{cos(2y)}$$

 

[mezarashi]

For the second step, I'd prefer having it written as:

$$-sin(y^2) \times 2y dy = 4x^3dx$$

This will probably be more consistent once you encounter more complicated problems or do multivariable calculus.

Answer B is correct. You forgot to move the y over in your last step, step 3.

 

[Doc Al]

Answer B is correct. You forgot to move the y over in your last step, step 3.

Right! (I just realized that you left out that y in your last step!)

 

[mlowery]

Hehe, how'd that "y" sneak by me?
Thanks for the help.

Yeah mezarashi, I am not quite familiar with the notation you used (I just started using the dy/dx notation last week).

Thanks,
Mitch