Find the derivative of the function: $\frac{x^{2} \cos \left(\frac{\pi}{4}\right)}{\sin x}$

Vedclass pdf generator app on play store
Vedclass iOS app on app store
Let $f(x) = \frac{x^{2} \cos \left(\frac{\pi}{4}\right)}{\sin x}$.
Using the quotient rule $\left(\frac{u}{v}\right)^{\prime} = \frac{u^{\prime}v - uv^{\prime}}{v^{2}}$,where $u = x^{2} \cos \left(\frac{\pi}{4}\right)$ and $v = \sin x$:
$f^{\prime}(x) = \cos \left(\frac{\pi}{4}\right) \left[ \frac{\sin x \cdot \frac{d}{dx}(x^{2}) - x^{2} \cdot \frac{d}{dx}(\sin x)}{\sin^{2} x} \right]$
$f^{\prime}(x) = \cos \left(\frac{\pi}{4}\right) \left[ \frac{\sin x(2x) - x^{2}(\cos x)}{\sin^{2} x} \right]$
$f^{\prime}(x) = \frac{x \cos \left(\frac{\pi}{4}\right) (2 \sin x - x \cos x)}{\sin^{2} x}$

Explore More

Similar Questions

Let $f(x) = x\sqrt{x\sqrt{x\sqrt{x\dots\infty}}}$ where $x > 0$. Then $f'(3)$ is equal to:

Difficult
View Solution

The first derivative of the function $\left[ \cos^{-1}\left( \sin \sqrt{\frac{1+x}{2}} \right) + x^x \right]$ with respect to $x$ at $x = 1$ is

If $y = \cos(\sin x^2)$,then $\frac{dy}{dx}$ at $x = \sqrt{\frac{\pi}{2}}$ is

Find the derivative: $\frac{d}{dx} \left( \frac{e^x}{1 + x^2} \right)$

Differentiate the function with respect to $x$: $\sin^{3} x + \cos^{6} x$

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