Easy
Differentiate the following with respect to $x:$ $\sin (\log x), x > 0$
Let $y = \sin (\log x).$
Using the chain rule,we have:
$\frac{dy}{dx} = \cos (\log x) \cdot \frac{d}{dx}(\log x)$
Since $\frac{d}{dx}(\log x) = \frac{1}{x},$ we get:
$\frac{dy}{dx} = \cos (\log x) \cdot \frac{1}{x} = \frac{\cos (\log x)}{x}$
Using the chain rule,we have:
$\frac{dy}{dx} = \cos (\log x) \cdot \frac{d}{dx}(\log x)$
Since $\frac{d}{dx}(\log x) = \frac{1}{x},$ we get:
$\frac{dy}{dx} = \cos (\log x) \cdot \frac{1}{x} = \frac{\cos (\log x)}{x}$
Explore More
Similar Questions
The graph of function $f$ contains the points $P(1, 2)$ and $Q(s, r)$. The equation of the secant line through $P$ and $Q$ is $y = \left( \frac{s^2 + 2s - 3}{s - 1} \right) x - 1 - s$. The value of $f'(1)$ is:
$x = \frac{1-\sqrt{y}}{1+\sqrt{y}} \Rightarrow \frac{dy}{dx}$ is equal to
For constant $a$,$\frac{d}{d x}\left(x^{x}+x^{a}+a^{x}+a^{a}\right)$ is
MediumKCET 2021
View SolutionIf $y=\log \left(\frac{1+x}{1-x}\right)^{1 / 4}-\frac{1}{2} \tan ^{-1}(x)$,then $\frac{d y}{d x}$ at $x=\frac{1}{\sqrt{2}}$ equals
Find the derivative of the following function (it is to be understood that $a, b, c$ are fixed non-zero constants): $\frac{1}{ax^{2}+bx+c}$
Medium
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo