Determine (dy/dx) using implicit differentiation

[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?

 

[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}

 

[koolkris623]

cool thanks

 

[cristo]

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.