Hey there mate!
Please check the attached answer of picture for explanation.
[tex]Have \: a \: great \: day! \: :)[/tex]
Answer:
[tex]\sf \dfrac{dy}{dx} =\sf \bold{ -nx\cot \left(x\right)\csc ^n\left(x\right) + \csc ^n\left(x\right)}[/tex]
solve:
[tex]\sf y = \dfrac{d}{dx}\left(\dfrac{x}{sin^nx}\right)[/tex]
[tex]\hookrightarrow \sf \bold{ \sf \dfrac{dy}{dx} =\sf \dfrac{d}{dx}\left(\dfrac{x}{sin^nx}\right)}[/tex]
// apply rule: [tex]\sf \dfrac{1}{sinx} = csc(x)[/tex] //
[tex]\hookrightarrow \sf \bold{ \sf \dfrac{dy}{dx} =\sf \dfrac{d}{dx}\left(x\csc ^n\left(x\right)\right)}[/tex]
// apply product rule: [tex]\sf xsinx = x * \frac{d}{dx} (sinx) + sin(x) *\frac{d}{dx} (x)[/tex] //
[tex]\hookrightarrow \sf \bold{\sf x *\frac{d}{dx} (csc^n (x))+ csc^n (x) * \dfrac{d}{dx} (x)}[/tex]
Lets look into deeper differentiation separately:
[tex]\sf we \ must \ know \ that \ \dfrac{d}{dx} (x)} = 1[/tex]
now, for [tex]\sf \frac{d}{dx} (csc^n (x))[/tex] - apply chain rule
[tex]\rightarrow \sf n\left(\csc \left(x\right)\right)^{n-1}\dfrac{d}{dx}\left(\csc \left(x\right)\right)[/tex]
[tex]\sf \bold \ * we \ must \ know \ that \ \dfrac{d}{dx} (csc(x)) = -cot(x) csc(x)[/tex]
[tex]\sf \rightarrow n\left(\csc \left(x\right)\right)^{n-1}\left(-\cot \left(x\right)\csc \left(x\right)\right)[/tex]
[tex]\rightarrow \sf -n\cot \left(x\right)\csc ^{n-1+1}\left(x\right)[/tex]
[tex]\rightarrow \sf -n\cot \left(x\right)\csc ^n\left(x\right)[/tex]
Now finish:
[tex]\hookrightarrow \sf x * -n\cot \left(x\right)\csc ^n\left(x\right) + \csc ^n\left(x\right) *1[/tex]
[tex]\hookrightarrow \sf \bold{ -nx\cot \left(x\right)\csc ^n\left(x\right) + \csc ^n\left(x\right)}[/tex]
