Let $f(x) = \sqrt{x - 1} + \sqrt{x + 24 - 10\sqrt{x - 1}}$ for $1 < x < 26$ be a real-valued function. Then $f'(x)$ for $1 < x < 26$ is

  • A
    $0$
  • B
    $\frac{1}{\sqrt{x - 1}}$
  • C
    $2\sqrt{x - 1} - 5$
  • D
    None of these

Explore More

Similar Questions

The values of $x$,at which the first derivative of the function $(\sqrt{x} + \frac{1}{\sqrt{x}})^2$ with respect to $x$ is $\frac{3}{4}$,are

If $x \neq 0$ and $f(x)$ satisfies $8 f(x) + 6 f(\frac{1}{x}) = x + 5$,then the value of $\frac{d}{dx} (x^2 f(x))$ at $x = 1$ is

If $y = \frac{x^{2/3} - x^{-1/3}}{x^{2/3} + x^{-1/3}}$,$x \neq 0$,then $(x+1)^2 y_1 = $

Find the derivative of the following function (it is to be understood that $p, q, r$ are fixed non-zero constants): $\frac{p x^{2}+q x+r}{x}$

For $x > 0$ and $(x \log x) < 1$, if $y = \cot^{-1} \left( \frac{x - \log x}{x^2 \log_e x^2 + \log x^x} \right)$, then $\frac{dy}{dx} = \dots$

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