Question

Find derivative of Y = x / sin n x​

Answer 1

here is your answer..,. attached.....

Answer 2

$\dfrac{dy}{dx}=\dfrac{\sin x-nx.\cos x}{\sin^{n+1} x}$

Step-by-step explanation:

We have,

$y=\dfrac{x}{\sin^n x}$

To find, derivative of $y=\dfrac{x}{\sin^n x}$

We know that,

$\dfrac{d(\dfrac{u}{v} )}{dx} =\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx} }{v^{2}}$

$\dfrac{dy}{dx} = \dfrac{d(\dfrac{x}{\sin^n x})}{dx}$

$=\dfrac{\sin^n x.1-x.n\sin^{n-1} x.\cos x}{(\sin^n x)^2}}$

$=\dfrac{\sin^n x.1-nx\sin^{n-1} x.\cos x}{(\sin^n x)^2}}$

$=\dfrac{\sin x-nx.\cos x}{\sin^{n+1} x}$

Hence,$\dfrac{dy}{dx}=\dfrac{\sin x-nx.\cos x}{\sin^{n+1} x}$