# Homework Help: Determine (dy/dx) using implicit differentiation

1. Oct 16, 2007

### koolkris623

Determine (dy/dx) using implicit differentiation.

cos(X^2Y^2) = x

I'm really confused what to do now..i think the next steps are:

d/dx [cos(X^2*Y^2)] = d/dx [x]
= -sin(X^2*Y^2)* ((X^2*2Y dy/dx) + (Y^2*2X)) = 1
= -2YX^2 sin(X^2*Y^2) dy/dx + -2XY^2sin(X^2*Y^2) = 1
= -2YX^2 sin(X^2*Y^2) dy/dx = 1 + -2XY^2sin(X^2*Y^2)
= dy/dx = (1 + -2XY^2sin(X^2*Y^2))/ (-2YX^2 sin(X^2*Y^2))

Can someone tell me if this is correct?

2. Oct 16, 2007

### atqamar

$$cos(x^2 y^2)=x$$

$$\frac{d}{dx}[cos(x^2 y^2)]=\frac{d}{dx}[x]$$

$$-sin(x^2 y^2) [\frac{d}{dx}(x^2 y^2)] = 1$$

$$-sin(x^2 y^2) [2y^2x + 2x^2y\frac{dy}{dx}] = 1$$

$$-sin(x^2 y^2)2y^2x -sin(x^2 y^2)2x^2y\frac{dy}{dx} = 1$$

$$-sin(x^2 y^2)2x^2y\frac{dy}{dx} = 1+sin(x^2 y^2)2y^2x$$

$$\frac{dy}{dx} = \frac{1+sin(x^2 y^2)2y^2x}{-sin(x^2 y^2)2x^2y}$$

3. Oct 16, 2007

### koolkris623

cool thanks

4. Oct 17, 2007

### cristo

Staff Emeritus
atqamar, please note that one should not provide full solutions in the homework forums, but rather should offer guidance and hints as to how to proceed.