Answer to Question #294739 in Calculus for ZARUL

Question

Find dy/dx of the function xy² + e^6x y³ = cos(x² + 2) at the point (1,1) by using implicit differentiation. 

Answer 1

Using implicit differentiation we get

$y^2+2xyy'+6e^{6x}y^3+e^{6x}3y^2y'=-\sin(x^2+2)2x.$

It follows that at the point $(1,1)$ we have

$1+2y'+6e^{6}+3e^{6}y'=-2\sin(3).$

Therefore,

$(2+3e^{6})y'=-1-6e^{6}-2\sin(3),$

and we conclude that

$\frac{dy}{dx}=y'=-\frac{1+6e^{6}+2\sin(3)}{2+3e^{6}}.$