Question

The derivative of cosec^-1 (x²) is​

Answer 1

Question :

Find the Derivatives of the $\sf\csc^{-1}(x^2)$

Formula's Used :

• Genral Formula

$1)\sf\:\frac{d(x {}^{n} )}{dx} = nx {}^{n - 1}$

$2)\sf\:\frac{d(constant)}{dx} = 0$

$3)\sf\dfrac{d(\cot\:x)}{dx}=\dfrac{-1}{1+x^2}$

$\sf4)\dfrac{d(\csc\:x)}{dx}=\dfrac{-1}{|x|\times\sqrt{x^2-1}}$

• Chain rule

Let y=f(t) ,t = g(u) and u =m(x) ,then

$\sf\:\dfrac{dy}{dx} = \dfrac{dy}{dt} \times \dfrac{dt}{du} \times \dfrac{du}{dx}$

Solution :

We have to find the derivative of$\sf\csc^{-1}(x^2)$

Let $\sf\:x^2=t$

Then,

$\sf\:y=\csc^{-1}(t)$

Now Differentiate it with respect to x, by chain rule

$\sf\implies\dfrac{dy}{dx}=\dfrac{\csc^{-1}(t)}{t}\times\dfrac{dt}{dx}$

$\sf\implies\dfrac{dy}{dx}=\dfrac{\csc^{-1}(t)}{t}\times\dfrac{d(x^2)}{dx}$

$\sf\implies\dfrac{dy}{dx}=\dfrac{-1}{|t|\times\sqrt{(t)^2-1}}\times\dfrac{d(x^2)}{dx}$

$\sf\implies\dfrac{dy}{dx}=\dfrac{-1}{|x^2|\times\sqrt{(x^2)^2-1}}\times\:2x$

$\sf\implies\dfrac{dy}{dx}=\dfrac{-2x}{|x^2|\times\sqrt{(x^2)^2-1}}$

$\sf\implies\dfrac{dy}{dx}=\dfrac{-2x}{|x^2|\times\sqrt{(x^4-1)}}$

Answer 2

Answer:

Answer is in this attachment.. Hope it's help you dear