Differentiate the function with respect to $x$: $\cos (\sin x)$

Vedclass pdf generator app on play store
Vedclass iOS app on app store
Let $f(x) = \cos (\sin x)$.
Using the chain rule,we differentiate with respect to $x$:
$\frac{d}{dx}[\cos (\sin x)] = -\sin (\sin x) \cdot \frac{d}{dx}(\sin x)$
Since $\frac{d}{dx}(\sin x) = \cos x$,we substitute this back:
$= -\sin (\sin x) \cdot \cos x$
$= -\cos x \sin (\sin x)$

Explore More

Similar Questions

For some constants $a$ and $b$,find the derivative of $(ax^2 + b)^2$.

If $f^{\prime}(x) = \tan^{-1}(\sec x + \tan x)$,$\frac{-\pi}{2} < x < \frac{\pi}{2}$ and $f(0) = 0$,then $f(1) =$

Differentiating the following with respect to $x$: $\frac{\cos x}{\log x}, x > 0$

If $y = \frac{e^x + e^{-x}}{e^x - e^{-x}}$,then $\frac{dy}{dx}$ is equal to

The derivative of $e^{3x} \sin 4x$ with respect to $x$ 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