Question

Use implicit differentiation on xy=ln(xcoty) to show dy/dx=((1-xy)coty)/(x^2coty+xcsc^2y)

Answer 1

\[{d \over dx}\left[ xy \right]={d \over dx}\left[\ln(x \cot y) \right]\]

Answer 2

\[y+x{dy \over dx}={1 \over x \cot y}\cdot \left( \cot y+ x \left( -\csc^2y \right){dy \over dx}\right)\]

Answer 3

multiply \(x\cot y\) to both sides:\[\implies xy \cot y+x^2\cot y{dy \over dx}=\cot y- x \csc^2y{dy \over dx}\]

\[\implies x^2\cot y{dy \over dx}+x \csc^2y{dy \over dx}=\cot y- xy \cot y\]\[\implies \left( x^2\cot y+x \csc^2y \right){dy \over dx}=\left( 1- xy \right) \cot y\]
\[\implies{dy \over dx}={\left( 1- xy \right) \cot y \over x^2\cot y+x \csc^2y}\]