Question

derivative of cosec-¹ 1+x²/2x​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{y=cosec^{-1}\bigg(\dfrac{1+x^2}{2x}\bigg)}$

$\tt{\implies\,y=sin^{-1}\bigg(\dfrac{2x}{1+x^2}\bigg)}$

$\sf{Let\,\,\,x=tan(\theta)}$

$\sf{\dfrac{dx}{d\theta}=sec^2(\theta)\,\,\,\,\,\,\,\,\,\,...(1)}$

Now,

$\tt{\implies\,y=sin^{-1}\bigg(\dfrac{2\,tan(\theta)}{1+tan^2(\theta)}\bigg)}$

$\tt{\implies\,y=sin^{-1}(sin(2\theta))}$

$\tt{\implies\,y=2\theta}$

$\tt{\implies\,\dfrac{dy}{d\theta}=2\,\,\,\,\,\,\,\,\,\,...(2)}$

Now, we have,

$\sf{\dfrac{dy}{dx}=\dfrac{\dfrac{dy}{d\theta}}{\dfrac{dx}{d\theta}}=\dfrac{2}{sec^2(\theta)}}$

$\sf{\implies\dfrac{dy}{dx}=\dfrac{2}{1+tan^2(\theta)}}$

$\sf{\implies\dfrac{dy}{dx}=\dfrac{2}{1+x^2}}$