Differentiate the function with respect to $x$: $\sin (ax+b)$

Vedclass pdf generator app on play store
Vedclass iOS app on app store
Let $f(x) = \sin (ax+b)$.
Using the chain rule,we differentiate with respect to $x$:
$\frac{d}{dx}[\sin (ax+b)] = \cos (ax+b) \cdot \frac{d}{dx}(ax+b)$
$= \cos (ax+b) \cdot [\frac{d}{dx}(ax) + \frac{d}{dx}(b)]$
$= \cos (ax+b) \cdot (a + 0)$
$= a \cos (ax+b)$

Explore More

Similar Questions

Let $f(x)$ be a differentiable function at $x=a$ with $f^{\prime}(a)=2$ and $f(a)=4$. Then $\lim _{x \rightarrow a} \frac{x f(a)-a f(x)}{x-a}$ equals ...... .

For constant $a$,$\frac{d}{d x}\left(x^{x}+x^{a}+a^{x}+a^{a}\right)$ is

The value of $\frac{d}{dx}[|x - 1| + |x - 5|]$ at $x = 3$ is

If $f(x) = 2x^2 + 3x - 5$,then the value of $f'(0) + 3f'(-1)$ is equal to

$A$ possible positive value of '$a$',for which $f^{\prime}(x)=0$ has equal roots,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