What is the implicit derivative of #1= xye^y-xcos(xy) #?

Question

What is the implicit derivative of #1=   xye^y-xcos(xy)    #?

Ezra Adams · Accepted Answer

#(dy)/(dx) = (cos(xy) - xysin(xy) - ye^y)/(xe^y+xye^y+x^2sin(xy))# #d/(dx)(1) = d/(dx)(xye^y) - d/(dx)(xcos(xy))# To do these derivatives, remember #d/(dx)(y) = (dy)/(dx)# and hammer your product and chain rules. #0 = ye^y + x(dy)/(dx)e^y + xy(dy)/(dx)e^y - cos(xy) + x(y+x(dy)/(dx))sin(xy)# #(dy)/(dx)(xe^y+xye^y+x^2sin(xy)) = cos(xy) - xysin(xy) - ye^y# #(dy)/(dx) = (cos(xy) - xysin(xy) - ye^y)/(xe^y+xye^y+x^2sin(xy))#

Charles Arocha · Answer

undefined To find the implicit derivative of $1 = xye^y - x\cos(xy)$ with respect to $x$, differentiate both sides of the equation with respect to $x$.

The derivative of $1$ with respect to $x$ is $0$.

For the term $xye^y$:
$$
\frac{d}{dx}(xye^y) = y \cdot e^y + xy \cdot \frac{d}{dx}(e^y) = y \cdot e^y + xy \cdot e^y \cdot \frac{dy}{dx}
$$

For the term $x\cos(xy)$:
$$
\frac{d}{dx}(x\cos(xy)) = \cos(xy) + x(-y\sin(xy)) = \cos(xy) - xy\sin(xy)
$$

Combine these results:
$$
0 = y \cdot e^y + xy \cdot e^y \cdot \frac{dy}{dx} - (\cos(xy) - xy\sin(xy))
$$

Solve for $\frac{dy}{dx}$:
$$
\frac{dy}{dx} = \frac{\cos(xy) - xy\sin(xy) - y \cdot e^y}{xy \cdot e^y}
$$