If f(x) = frac{x}{1+|x|},for x in mathbb{R},t | vedclass

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

Question

(B) Given $f(x) = \frac{x}{1+|x|}$.To find $f^{\prime}(0)$,we use the definition of the derivative:$f^{\prime}(0) = \lim_{h 	o 0} \frac{f(0+h) - f(0)

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) Given $f(x) = \frac{x}{1+|x|}$.To find $f^{\prime}(0)$,we use the definition of the derivative:$f^{\prime}(0) = \lim_{h 	o 0} \frac{f(0+h) - f(0)}{h}$.Since $f(0) = \frac{0}{1+|0|} = 0$,we have:$f^{\prime}(0) = \lim_{h 	o 0} \frac{\frac{h}{1+|h|} - 0}{h} = \lim_{h 	o 0} \frac{1}{1+|h|}$.Now,we evaluate the left-hand and right-hand limits:Left-hand limit $(h 	o 0^-)$: $\lim_{h 	o 0^-} \frac{1}{1-h} = 1$.Right-hand limit $(h 	o 0^+)$: $\lim_{h 	o 0^+} \frac{1}{1+h} = 1$.Since both limits are equal,$f^{\prime}(0) = 1$.

If $f(x) = \frac{x}{1+|x|}$,for $x \in \mathbb{R}$,then $f^{\prime}(0)$ is equal to :

Similar Questions

If $f(x) = \frac{\sin^2 x}{1+\cot x} + \frac{\cos^2 x}{1+\tan x}$,then the value of $f^{\prime}\left(\frac{\pi}{6}\right)$ is equal to

If $y=(1+x)(1+x^2)(1+x^4) \dots (1+x^{2^n})$, then $\left(\frac{dy}{dx}\right)_{x=0}$ is equal to

If $y=(1+x)(1+x^{2})(1+x^{4}) \ldots (1+x^{2^{n}}),$ then the value of $\left(\frac{d y}{d x}\right)$ at $x=0$ is

Find the derivative of $\tan (2 x+3)$

If $y = \log \tan \sqrt{x}$,then the value of $\frac{dy}{dx}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module