Question

Please solve this i will mark u as brainlist​

Answer 1

Step-by-step explanation:

We have,

$y = \ln( \tan(x) )$

Differentiating both sides w.r.t. x, we get,

$\implies \: \frac{dy}{dx} = \frac{ \sec^{2} (x) }{ \tan(x) }$

$\implies \: \frac{dy}{dx} = \frac{ \cos(x) }{ \cos^{2}(x) \sin(x) }$

$\implies \: \frac{dy}{dx} = \frac{ 1 }{ \cos(x) \sin(x) }$

$\implies \: \frac{dy}{dx} = \frac{ 2 }{ 2\sin(x) \cos(x) }$

$\implies \: \frac{dy}{dx} = \frac{ 2 }{ \sin(2x) }$

$\implies \: \frac{dy}{dx} = 2 \cosec(2x)$

Again, differentiating w.r.t x ,

$\implies \: \frac{d^{2} y}{dx^{2} } = - 4 \cosec(2x) \cot(2x)$

$\implies \: \frac{d^{2} y}{dx^{2} } = - \frac{2}{ \sin(2x) } .2 \cot(2x)$

$\implies \: \sin(2x) \frac{d^{2} y}{dx^{2} } = - 4 \cot(2x)$