Question

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

Answer 1

\[\sec^2(x-y)(1-y')=\frac{(2+x^2)y'-y(2x)}{(2+x^2)^2}\] is a start

Answer 2

then a raft of not very interesting algebra to solve this equation for \(y'\)

Answer 3

that's the part that's tripping me up, the algebra

Answer 4

how do i go about getting y' on one side

Answer 5

just gotta grind it out

Answer 6

distribute first

Answer 7

haha don't know where to start

Answer 8

\[\sec^2(x-y)(1-y')=\frac{(2+x^2)y'-y(2x)}{(2+x^2)^2}\]
\[\sec^2(x-y)-y'\sec^2(x-y)=\frac{(2+x^2)y'-2xy)}{(2+x^2)^2}\]

Answer 9

\[\sec^2(x-y)-y'\sec^2(x-y)=\frac{y'}{2+x^2}-\frac{2xy}{(2+x^2)^2}\]

Answer 10

\[\sec^2(x-y)+\frac{2xy}{(2+x^2)^2}=\frac{y'}{2+x^2}+y'\sec^2(x-y)\]

Answer 11

\[\sec^2(x-y)+\frac{2xy}{(2+x^2)^2}=y'\left(\frac{1}{2+x^2}+\sec^2(x-y)\right)\]

Answer 12

thanks! this helped