Question

differentiate using implicit diff tan^3 (xy^2+y)=x

Answer 1

is this same question?

Answer 2

yea i forgot the brackets!

Answer 3

oh so my solution is completly wrong then, haha :)

Answer 4

pliz try to solve it again

Answer 5

you have to use chain rule

\[(xy^{2} +y)' = y^{2} + 2xy \frac{dy}{dx} + \frac{dy}{dx}\]
\[\rightarrow 3\tan^{2}(xy^{2}+y) \sec^{2}(xy^{2}+y)(y^{2} + 2xy \frac{dy}{dx} + \frac{dy}{dx}) = 1\]
For simplicity, let u = tan^2*sec^2
\[\rightarrow 3uy^{2} +6uxy \frac{dy}{dx}+3u \frac{dy}{dx} = 1\]
\[\frac{dy}{dx} = \frac{1-3uy^{2}}{3u(2xy+1)}\]