(A) Let $y = e^{-x}$.Using the chain rule for differentiation,we have:$\frac{dy}{dx} = \frac{d}{dx}(e^{-x})$Since the derivative of $e^u$ with respect

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(A) Let $y = e^{-x}$.Using the chain rule for differentiation,we have:$\frac{dy}{dx} = \frac{d}{dx}(e^{-x})$Since the derivative of $e^u$ with respect to $u$ is $e^u$,we apply the chain rule:$\frac{dy}{dx} = e^{-x} \cdot \frac{d}{dx}(-x)$$\frac{dy}{dx} = e^{-x} \cdot (-1)$$\frac{dy}{dx} = -e^{-x}$

Class 12 Mathematics Continuity and Differentiation Derivative at a point, Standard differentiation

Easy

Differentiate the following with respect to $x$: $e^{-x}$

A
$-e^{-x}$
B
$e^{-x}$
C
$-x e^{-x-1}$
D
$e^{x}$

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Mathematics Questions

Chapter

Continuity and Differentiation

Subtopic

Derivative at a point, Standard differentiation

Find the derivative: $\frac{d}{dx} \left[ \log \left( x + \frac{1}{x} \right) \right] = $

Easy

View Solution

Differentiate the following with respect to $x$: $\sqrt{e^{\sqrt{x}}}, x > 0$

Medium

View Solution

The derivative of the function $\cot ^{-1}[\cos 2 x]^{1 / 2}$ at $x=\pi / 6$ is

MediumMHT CET 2021

View Solution

$\frac{d}{d x}\left[a \tan ^{-1} x+b \log \left(\frac{x-1}{x+1}\right)\right]=\frac{1}{x^4-1}$
$\Rightarrow a-2 b$ is equal to

MediumAP EAMCET 2009

View Solution

If $f: R \rightarrow R$ is a differentiable function at $a \in R$ such that $f^{\prime}(a)=a f(a)$, then $\lim _{x \rightarrow a} \frac{x f(a)-a f(x)}{x-a}=$

MediumTS EAMCET 2021

View Solution

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

Differentiate the following with respect to $x$: $e^{-x}$

Similar Questions

Find the derivative: $\frac{d}{dx} \left[ \log \left( x + \frac{1}{x} \right) \right] = $

Differentiate the following with respect to $x$: $\sqrt{e^{\sqrt{x}}}, x > 0$

The derivative of the function $\cot ^{-1}[\cos 2 x]^{1 / 2}$ at $x=\pi / 6$ is

$\frac{d}{d x}\left[a \tan ^{-1} x+b \log \left(\frac{x-1}{x+1}\right)\right]=\frac{1}{x^4-1}$ $\Rightarrow a-2 b$ is equal to

If $f: R \rightarrow R$ is a differentiable function at $a \in R$ such that $f^{\prime}(a)=a f(a)$, then $\lim _{x \rightarrow a} \frac{x f(a)-a f(x)}{x-a}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module

$\frac{d}{d x}\left[a \tan ^{-1} x+b \log \left(\frac{x-1}{x+1}\right)\right]=\frac{1}{x^4-1}$
$\Rightarrow a-2 b$ is equal to