Question

Differentate log(1+theta) with respect to sin inverse theta

Answer 1

Answer:

$\frac{ \sqrt{1 - { theta}^{2} } }{(1 + theta) ln(10) }$

Answer 2

$\frac{d(log(1+\theta)}{dsin^{-1}\theta}=\frac{\sqrt{1-\theta^2}}{1+\theta}$

Step-by-step explanation:

Let $t=sin^{-1}\theta$

$sint=\theta$

$y=log(1+\theta)=log(1+sint)$

Differentiate w.r.t t

$\frac{dy}{dt}=\frac{d(log(1+sint)}{dt}$

$\frac{dy}{dt}=\frac{1}{1+sint}(cost)$

Using the  formula

$\frac{d(logx)}{dx}=\frac{1}{x}$

$\frac{d(sinx)}{dx}=cosx$

$\frac{dy}{dt}=\frac{\sqrt{1-sin^2t}}{1+sint}=\frac{\sqrt{1-\theta^2}}{1+\theta}$

Using the formula

$cost=\sqrt{1-sin^2t}$

$\frac{d(log(1+\theta)}{dsin^{-1}\theta}=\frac{\sqrt{1-\theta^2}}{1+\theta}$

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