If g(x)=∫ limits sin x sin (2 x) sin -1(t) d t, then

Question

If g(x)=∫ limits sin x sin (2 x) sin -1(t) d t, then (A) g'((π/2))=-2 π (B) g'(-(π/2))=2 π (C) g'((π/2))=2 π (D) g'(-(π/2))=-2 π

Tardigrade Admin · Accepted Answer

Given, g(x)=∫ limits sin x sin (2 x) sin -1(t) d t Applying Newton Leibniz Formula, g prime(x)= sin -1[ sin (2 x)] × 2 cos 2 x- sin -1( sin x) × cos x ⇒ g prime(x)=4 x cos 2 x-x cos x So, g prime((π/2))=-2 π-0=-2 π And g prime(-(π/2))=2 π-0=2 π

Q. If $g (x) = s i n x \int s i n (2 x) sin^{- 1} (t) d t$ , then

$g^{'} (\frac{π}{2}) = - 2 π$

$g^{'} (- \frac{π}{2}) = 2 π$

$g^{'} (\frac{π}{2}) = 2 π$

$g^{'} (- \frac{π}{2}) = - 2 π$

Q. If g(x)=sinx∫sin(2x)​sin−1(t)dt, then

g′(2π​)=−2π

g′(−2π​)=2π

g′(2π​)=2π

g′(−2π​)=−2π

Given, g(x)=sinx∫sin(2x)​sin−1(t)dt Applying Newton Leibniz Formula, g′(x)=sin−1[sin(2x)]×2cos2x−sin−1(sinx)×cosx ⇒g′(x)=4xcos2x−xcosx So, g′(2π​)=−2π−0=−2π And g′(−2π​)=2π−0=2π

Q. If $g (x) = s i n x \int s i n (2 x) sin^{- 1} (t) d t$ , then

$g^{'} (\frac{π}{2}) = - 2 π$

$g^{'} (- \frac{π}{2}) = 2 π$

$g^{'} (\frac{π}{2}) = 2 π$

$g^{'} (- \frac{π}{2}) = - 2 π$