Question

If y = log(sin(logx)), find dy
/
dx. ?​

Answer 1

$\rm \longrightarrow y = log(sin(logx))$

$\rm \longrightarrow \dfrac{dy}{dx} = \dfrac{d \bigg(log(sin(logx)) \bigg)}{dx}$

$\rm \longrightarrow \dfrac{dy}{dx} = \dfrac{1}{sin(logx)} \times \dfrac{d \bigg(sin(logx) \bigg)}{dx}$

$\rm \longrightarrow \dfrac{dy}{dx} = \dfrac{1}{sin(logx)} \times cos(logx) \times \dfrac{d \bigg(logx \bigg)}{dx}$

$\rm \longrightarrow \dfrac{dy}{dx} = \dfrac{1}{sin(logx)} \times cos(logx) \times \dfrac{1}{x} \times \dfrac{d (x )}{dx}$

$\rm \longrightarrow \dfrac{dy}{dx} = \dfrac{1}{sin(logx)} \times cos(logx) \times \dfrac{1}{x} \times 1$

$\rm \longrightarrow \dfrac{dy}{dx} = \dfrac{cos(logx)}{sin(logx)} \times \dfrac{1}{x}$

$\rm \longrightarrow \dfrac{dy}{dx} = \dfrac{cos(logx)}{x \: sin(logx)}$

$\rm \longrightarrow \dfrac{dy}{dx} = \dfrac{cot(logx)}{x}$

Learn more :-

d(e^x)/dx = e^x

d(x^n)/dx = n x^(n-1)

d(ln x)/dx = 1/x

d(sin x)/dx = cos x

d(cos x)/dx = - sin x

d(tan x)/dx = sec² x

d(sec x)/dx = tan x * sec x

d(cot x)/dx = - cosec²x

d(cosec x)/dx = - cosec x * cot x

Answer 2

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