You visited us 1 times! Enjoying our articles? Unlock Full Access!

Chain Rule of Differentiation

A function ...

Question

A function $f$ , defined for all positive real numbers, satisfies the equation $f (x^{2}) = x^{3}$ fro every $x > 0$ . Then the value of $f^{'} (4) =$

12

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

3

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

3 / 2

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Cannot be determined

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct options are
A $12$
B $3$
C $3 / 2$
D Cannot be determined
Given, $f (x^{2}) = x^{3}, x > 0$
Differentiate on both sides w.r..t. x
$2 x f^{'} (x^{2}) = 3 x^{2}$
$∴ f^{'} (x^{2}) = \frac{3 x}{2}$
For $x = 2, f^{'} (4) = 3$

Suggest Corrections

Similar questions

f (x)

x > 0

f (x^{2}) = x^{3}

x > 0

f^{'} (4)

f

x

a

f (x + a) = \frac{1}{2} + \sqrt{f (x) - (f (x))^{2}}

x

1

0

f : R \to R

f (x^{2}) . f^{''} (x) = f^{'} (x) . f^{'} (x^{2})

x

f (1) = 1

f^{'} (1) + f^{''} (1)

f (x)

x

(x^{2} + x (f (x) - 2) + 2 \sqrt{3} - 3 - \sqrt{3} f (x)) = 0 \forall x \in R

f (\sqrt{3})

f

x

f (x) + 2 f (\frac{1}{x}) = 3 x .

x

f (x) = f (- x)

Join BYJU'S Learning Program