Let g(x)=x3 ln (x2 f(x)) where f(x) is a differentiable positive function on (0, ∞) satisfying f(2)=(1/4) and f prime(2)=-3, then g prime(2) equals

Question

Let g(x)=x3 ln (x2 f(x)) where f(x) is a differentiable positive function on (0, ∞) satisfying f(2)=(1/4) and f prime(2)=-3, then g prime(2) equals (A) 77 (B) -77 (C) 88 (D) -88

Tardigrade Admin · Accepted Answer

imodsC We have g(x)=x3 ln (x2 f(x)) ∴ On differentiating b.t.s. w.r.t. x, we get g prime(x)=(x3/x2 f(x))(x2 f prime(x)+2 x f(x))+3 x2 ln (x2 f(x)) text Hence g prime(2)=(2/f(2))(4 f prime(2)+4 f(2))+12 ln (4 f(2)) =8(4 ×-3+4 × (1/4))+12 ln (4 × (1/4))=8(-12+1)=-88

Q. Let g(x)=x3ln(x2f(x)) where f(x) is a differentiable positive function on (0,∞) satisfying f(2)=41​ and f′(2)=−3, then g′(2) equals

77

-77

88

-88

imodsC We have g(x)=x3ln(x2f(x)) ∴ On differentiating b.t.s. w.r.t. x, we get g′(x)=x2f(x)x3​(x2f′(x)+2xf(x))+3x2ln(x2f(x)) Hence g′(2)=f(2)2​(4f′(2)+4f(2))+12ln(4f(2)) =8(4×−3+4×41​)+12ln(4×41​)=8(−12+1)=−88

Q. Let $g (x) = x^{3} ln (x^{2} f (x))$ where $f (x)$ is a differentiable positive function on $(0, \infty)$ satisfying $f (2) = \frac{1}{4}$ and $f^{'} (2) = - 3$ , then $g^{'} (2)$ equals