Let g(x)=f(x) sin x, where f(x) is a twice differentiable function on (-∞, ∞) such that f prime(-π)=1. The value of g prime prime(-π) equals

Question

Let g(x)=f(x) sin x, where f(x) is a twice differentiable function on (-∞, ∞) such that f prime(-π)=1. The value of g prime prime(-π) equals (A) 1 (B) 2 (C) -2 (D) 0

Tardigrade Admin · Accepted Answer

We have g ( x )= f ( x ) sin x ....(1) On differentiating equation (1) w.r.t. x, we get g prime( x )= f ( x ) cos x + f prime( x ) sin x ....(2) Again differentiating equation (2) w.r.t. x, we get g prime prime(x)=f(x)(- sin x)+f prime(x) cos x+f prime(x) cos x+f prime prime(x) sin x ......(3) ⇒ g prime prime(-π)=2 f prime(-π) cos (-π)=2 × 1 ×-1=-2 text Hence g prime prime(-π)=-2

Q. Let g(x)=f(x)sinx, where f(x) is a twice differentiable function on (−∞,∞) such that f′(−π)=1. The value of g′′(−π) equals

1

2

-2

0

Q. Let $g (x) = f (x) sin x$ , where $f (x)$ is a twice differentiable function on $(- \infty, \infty)$ such that $f^{'} (- π) = 1$ . The value of $g^{''} (- π)$ equals