If g(x) = \int_{sinx}^{sin 2 x} \left(sin\right)^{- 1} \left(\right. t \left.\right) \textrm{ } dt , then

Engineering

Mathematics

Leibnitz Rule Derivative of Anti Derivatives

Question

If g(x) = $\int_{sinx}^{\sin 2 x} \sin^{- 1} (t) dt$ , then

$g' (\frac{- π}{2}) = 2 π$

$g' (\frac{π}{2}) = 2 π$

$g' (\frac{- π}{2}) = - 2 π$

$g' (\frac{π}{2}) = - 2 π$

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$g (x) = \int_{sinx}^{\sin (2 x)} \sin^{- 1} (t) dt$

g'(x) = 2 (cos 2x) sin^–1 (sin 2x) – (cos x) sin^–1(sin x)

$g' (\frac{π}{2}) = 2 (- 1) (0) - \cos (\frac{π}{2}) (1) = 0$

$g' (\frac{π}{2}) = 0$

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