Question

find the derivative of x^2 cosec^-1 (1/lnx)​

Answer 1

Step-by-step explanation:

We have,

$y = {x}^{2} \cosec^{ - 1} \bigg( \frac{1}{ \ln(x) } \bigg)$

$\implies \: y = {x}^{2} \sin^{ - 1} ( \ln(x) )$

$\implies \: \frac{dy}{dx} = 2x \sin^{ - 1} ( \ln(x) ) + {x}^{2}. \frac{1}{ \sqrt{1 - ( \ln(x))^{2} } } . \frac{d}{dx}( \ln(x))$

$\implies \: \frac{dy}{dx} = 2x \sin^{ - 1} ( \ln(x) ) + {x}^{2}. \frac{1}{ \sqrt{1 - ( \ln(x))^{2} } } . \frac{1}{x}$

$\implies \: \frac{dy}{dx} = 2x \sin^{ - 1} ( \ln(x) ) + x. \frac{1}{ \sqrt{1 - ( \ln(x))^{2} } }$

Hence,

$\tt \red{ \frac{dy}{dx} = 2x \sin^{ - 1} ( \ln(x) ) + \frac{x}{ \sqrt{1 - ( \ln(x))^{2} } } }$