If `g(x)=int_(sinx)^("sin"(2x))sin^(-1)(t)dt ,t h e n :`
(a) `g^(prime)(pi/2)=-2pi` (b) `g^(prime)(-pi/2)=-2pi` (c) `g^(prime)(-pi/2)=2pi` (d) `g^(prime)(pi/2)=2pi`

Options

A. `g'((pi)/2)=-2pi`
B. `g'(-(pi)/2)=2pi`
C. `g'((pi)/2)=2pi`
D. `g'(-(pi)/2)=-2pi`

Explanation

To solve the problem, we need to differentiate the function \( g(x) \) defined as:

\[ g(x) = \int_{\sin x}^{\sin(2x)} \sin^{-1}(t) \, dt \]

We will use the Leibniz rule for differentiation under the integral sign.

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