If $f(x)$ and $g(x)$ are two functions with $g(x)=x-\frac{1}{x}$ and $f \circ g(x)=x^3-\frac{1}{x^3}$,then $f^{\prime}(x)$ is equal to

  • A
    $3x^2+\frac{3}{x^4}$
  • B
    $x^2-\frac{1}{x^2}$
  • C
    $1-\frac{1}{x^2}$
  • D
    $3x^2+3$

Explore More

Similar Questions

Let $f: R \rightarrow R$ be defined as $f(x) = 2x - 1$ and $g: R - \{1\} \rightarrow R$ be defined as $g(x) = \frac{x - 1/2}{x - 1}$. Then the composition function $f(g(x))$ is:

Let $f: \{1,3,4\} \rightarrow \{1,2,5\}$ and $g: \{1,2,5\} \rightarrow \{1,3\}$ be given by $f = \{(1,2), (3,5), (4,1)\}$ and $g = \{(1,3), (2,3), (5,1)\}$. Write down $g \circ f$.

If $f: R \rightarrow R$ and $g: R^{+} \rightarrow R$ are such that $g\{f(x)\}=|\sin x|$ and $f\{g(x)\}=(\sin \sqrt{x})^2$, then a possible choice for $f$ and $g$ is

Consider the function $f: R \rightarrow R$ defined by $f(x)=\frac{2x}{\sqrt{1+9x^2}}$. If the composition of $f$,$\underbrace{(f \circ f \circ \ldots \circ f)}_{10 \text{ times }}(x) = \frac{2^{10}x}{\sqrt{1+9\alpha x^2}}$,then the value of $\sqrt{3\alpha+1}$ is equal to:

Let $f: R \rightarrow R$ be a function defined by $f(x) = \left(2\left(1 - \frac{x^{25}}{2}\right)\left(2 + x^{25}\right)\right)^{\frac{1}{50}}$. If the function $g(x) = f(f(f(x))) + f(f(x))$,then the greatest integer less than or equal to $g(1)$ is

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo