Let f(x + y) = f(x) + f(y) and f(x) = x^2 g(x | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) We know that the derivative is defined as $f'(x) = \lim_{h 	o 0} \frac{f(x + h) - f(x)}{h}$.Since $f(x + y) = f(x) + f(y)$,we have $f(x + h) = f(

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(D) We know that the derivative is defined as $f'(x) = \lim_{h 	o 0} \frac{f(x + h) - f(x)}{h}$.Since $f(x + y) = f(x) + f(y)$,we have $f(x + h) = f(x) + f(h)$.Substituting this into the derivative formula,we get $f'(x) = \lim_{h 	o 0} \frac{f(x) + f(h) - f(x)}{h} = \lim_{h 	o 0} \frac{f(h)}{h}$.Given $f(x) = x^2 g(x)$,we have $f(h) = h^2 g(h)$.Thus,$f'(x) = \lim_{h 	o 0} \frac{h^2 g(h)}{h} = \lim_{h 	o 0} h \cdot g(h)$.Since $g(x)$ is a continuous function,$\lim_{h 	o 0} g(h) = g(0)$.Therefore,$f'(x) = 0 \cdot g(0) = 0$.

Let $f(x + y) = f(x) + f(y)$ and $f(x) = x^2 g(x)$ for all $x, y \in R$,where $g(x)$ is a continuous function. Then $f'(x)$ is equal to

Similar Questions

$\frac{d}{d x}\left(e^{\log _e \sqrt{1+\tan ^2 x}}\right) =$

Differentiate the following with respect to $x$: $\frac{e^{x}}{\sin x}$

The differential coefficient of $\sqrt{\sec \sqrt{x}}$ is

Differentiate the following with respect to $x$: $e^{-x}$

If $y = x \cdot 7^x$, then the value of $\frac{dx}{dy}$ when $x = 1$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module