Answer to Question #137413 in Calculus for Nick

Question

use implicit differentiantion to find the derivative of the fuction

cos(x^2+2y) + xe^y = 0

Answer 1

$cos(x^2+2y) + xe^y = 0$

$\frac{d}{dx}[cos(x^2+2y) + xe^y]=\frac{d}{dx}[0]$

$\frac{d}{dx}[cos(x^2+2y) + xe^y]=\frac{d}{dx}[0]$

$(-sin(x^2+2y)).\frac{d}{dx}[x^2+2y]+\frac{d}{dx}[e^y].x+e^y.\frac{d}{dx}[x]=0$

$-(\frac{d}{dx}[x^2]+\frac{d}{dx}[2y])sin(x^2+2y)+e^y.\frac{d}{dx}[y].x+e^y.1=0$

$-(2x+2y\prime)sin(x^2+2y)+e^yy\prime x+e^y=0$

$-(2y\prime +2x)sin(2y+x^2)+xe^yy\prime +e^y=0$

$y\prime =\frac{2xsin(2y+x^2)-e^y}{2sin(2y+x^2)-xe^y}$