If f(x) = frac{x}{1+x} and g(x) = f(f(x)),the | vedclass

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

Question

(D) Given,$f(x) = \frac{x}{1+x}$.We need to find $g(x) = f(f(x))$.$g(x) = f\left(\frac{x}{1+x}\right) = \frac{\frac{x}{1+x}}{1 + \frac{x}{1+x}}$.Multi

Vedclass · Accepted Answer

$\frac{1}{(2x+1)^2}$

Vedclass · Answer

(D) Given,$f(x) = \frac{x}{1+x}$.We need to find $g(x) = f(f(x))$.$g(x) = f\left(\frac{x}{1+x}\right) = \frac{\frac{x}{1+x}}{1 + \frac{x}{1+x}}$.Multiplying the numerator and denominator by $(1+x)$,we get:$g(x) = \frac{x}{1+x+x} = \frac{x}{2x+1}$.Now,differentiate $g(x)$ with respect to $x$ using the quotient rule $\left(\frac{u}{v}\right)^{\prime} = \frac{u^{\prime}v - uv^{\prime}}{v^2}$:$g^{\prime}(x) = \frac{(1)(2x+1) - (x)(2)}{(2x+1)^2}$.$g^{\prime}(x) = \frac{2x+1 - 2x}{(2x+1)^2} = \frac{1}{(2x+1)^2}$.

If $f(x) = \frac{x}{1+x}$ and $g(x) = f(f(x))$,then $g^{\prime}(x)$ is equal to

Similar Questions

The derivative of $y = (1-x)(2-x)(3-x) \dots (n-x)$ at $x=1$ is

Find the value of $k$ if $\frac{d}{d x}\left\{\frac{2}{\sqrt{2+\sqrt{2+\sqrt{2+2 \cos (4 x)}}}}\right\} = k \sec \left(\frac{x}{2}\right) \tan \left(\frac{x}{2}\right)$

Let $f$ be a differentiable function such that $f(1) = 2$ and $f'(x) = f(x)$ for all $x \in R$. If $h(x) = f(f(x))$,then $h'(1)$ is equal to

If $y = e^{\sqrt{x}} + e^{-\sqrt{x}}$,then $\frac{dy}{dx}$ equals:

Let $m$ and $n$ be odd integers such that $0 < m < n$. If $f(x) = x^{\frac{m}{n}}$ for $x \in \mathbb{R}$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module