Question

Find dy/dx by implicit differentiation
tan(x-y)=y/(5+x^2)

Answer 1

hello are you there?

Answer 2

Yech.. tan.

\[\frac{d}{dx}[tan(x-y)] \]\[= sec^2(x-y)\frac{d}{dx}(x-y)\]\[=sec^2(x-y)(1 - \frac{dy}{dx})\]

\[\frac{d}{dx}[{y\over5+x^2}] \]\[= \frac{d}{dx}[y(5+x^2)^{-1}]\]\[= y\frac{d}{dx}[(5+x^2)^{-1}] + \frac{1}{5+x^2}\frac{dy}{dx}\]\[= -y(5+x^2)^{-2}(2x) + \frac{1}{5+x^2}\frac{dy}{dx}\]\[=\frac{1}{5+x^2}\frac{dy}{dx} - \frac{2xy}{(5+x^2)^2}\]