Easy
Differentiate the function with respect to $x$: $\sin(x^{2}+5)$.
Let $f(x) = \sin(x^{2}+5)$.
Using the chain rule,we differentiate the outer function $\sin(u)$ and multiply it by the derivative of the inner function $u = x^{2}+5$.
$\frac{d}{dx}[\sin(x^{2}+5)] = \cos(x^{2}+5) \cdot \frac{d}{dx}(x^{2}+5)$.
Since $\frac{d}{dx}(x^{2}) = 2x$ and $\frac{d}{dx}(5) = 0$,we have $\frac{d}{dx}(x^{2}+5) = 2x$.
Therefore,$\frac{d}{dx}[\sin(x^{2}+5)] = \cos(x^{2}+5) \cdot 2x = 2x \cos(x^{2}+5)$.
Using the chain rule,we differentiate the outer function $\sin(u)$ and multiply it by the derivative of the inner function $u = x^{2}+5$.
$\frac{d}{dx}[\sin(x^{2}+5)] = \cos(x^{2}+5) \cdot \frac{d}{dx}(x^{2}+5)$.
Since $\frac{d}{dx}(x^{2}) = 2x$ and $\frac{d}{dx}(5) = 0$,we have $\frac{d}{dx}(x^{2}+5) = 2x$.
Therefore,$\frac{d}{dx}[\sin(x^{2}+5)] = \cos(x^{2}+5) \cdot 2x = 2x \cos(x^{2}+5)$.
Explore More
Similar Questions
Differentiate the function with respect to $x$: $\sin^{-1}(x\sqrt{x})$,where $0 \le x \le 1$.
Medium
View SolutionIf $y = x \left[ \left( \cos \frac{x}{2} + \sin \frac{x}{2} \right) \left( \cos \frac{x}{2} - \sin \frac{x}{2} \right) + \sin x \right] + \frac{1}{2\sqrt{x}}$,then $\frac{dy}{dx} = $
Medium
View SolutionIf $f:R \to R$ is a differentiable function and $f(1) = 4$,then the value of $\lim_{x \to 1} \int_4^{f(x)} \frac{2t}{x - 1} dt$ is:
Difficult
View SolutionIf $f(x) = \begin{cases} x-5, & \text{for } x \leq 1 \\ 4x^2-9, & \text{for } 1 < x < 2 \\ 3x+4, & \text{for } x \geq 2 \end{cases}$,then $f^{\prime}(2^{+})$ is equal to:
$\frac{d}{dx} [\operatorname{cosech}^{-1}(\tan 2x)] = $
Vedclass 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