A differentiable function $f(x)$ is defined for all $x 0$ and satisfies $f\left( x^{3} \right) = 4x^{4}$ for all $x 0.$ The value of $f^{'}(8)$ is

Question

A differentiable function $f(x)$ is defined for all $x > 0$ and satisfies $f\left( x^{3} \right) = 4x^{4}$ for all $x > 0.$ The value of $f^{'}(8)$ is

Dr. Abubacker · Accepted Answer

(b)
$f\left( x^{3} ight) = 4x^{4},\forall x > 0$
Let $x^{3} = t\ \ \  \Rightarrow$ $x = t^{1/3}$
$	herefore\ f(t) = 4t^{4/3}\$
On differentiating w.r.t. $t,\$we get
$f^{'}(t) = 4.\frac{4}{3}(t)^{1/3}$
$	herefore\ f^{'}\left( x^{3} ight) = \frac{16}{3}\left( x^{3} ight)^{1/3} = \frac{16}{3}x$
$	herefore{\ f}^{'}(8) = f^{'}\left( 2^{3} ight) = \frac{16}{3} 	imes 2 = \frac{32}{3}$

Q. A differentiable function $f(x)$ is defined for all $x > 0$ and satisfies $f\left( x^{3} \right) = 4x^{4}$ for all $x > 0.$ The value of $f^{'}(8)$ is

Explanation

Reviewed by Dr. Abubacker

Want more practice?