The differential of f(x) = log_{e}(1 + e^{10x | vedclass

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

Question

(A) The differential $df$ is given by $df = f'(x) dx$. First,we find the derivative $f'(x)$: $f'(x) = \frac{d}{dx} [\log_{e}(1 + e^{10x}) - 	an^{-1}(

Vedclass · Accepted Answer

$0.5$

Vedclass · Answer

(A) The differential $df$ is given by $df = f'(x) dx$. First,we find the derivative $f'(x)$: $f'(x) = \frac{d}{dx} [\log_{e}(1 + e^{10x}) - 	an^{-1}(e^{5x})]$$f'(x) = \frac{1}{1 + e^{10x}} \cdot (10e^{10x}) - \frac{1}{1 + (e^{5x})^2} \cdot (5e^{5x})$$f'(x) = \frac{10e^{10x}}{1 + e^{10x}} - \frac{5e^{5x}}{1 + e^{10x}}$At $x = 0$,$e^{10(0)} = e^0 = 1$ and $e^{5(0)} = e^0 = 1$. $f'(0) = \frac{10(1)}{1 + 1} - \frac{5(1)}{1 + 1} = \frac{10}{2} - \frac{5}{2} = \frac{5}{2} = 2.5$.Now,calculate the differential for $dx = 0.2$: $df = f'(0) \cdot dx = 2.5 \cdot 0.2 = 0.5$.

The differential of $f(x) = \log_{e}(1 + e^{10x}) - \tan^{-1}(e^{5x})$ at $x = 0$ and for $dx = 0.2$ is

Similar Questions

The differential coefficient of $a^x + \log x \cdot \sin x$ is

Find the derivative of $\frac{x^{n}-a^{n}}{x-a}$ with respect to $x$,where $a$ is a constant.

$\frac{d}{dx}(e^{x^3})$ is equal to

If the function $f(x)$ is defined by $f(x) = \frac{x^{100}}{100} + \frac{x^{99}}{99} + \dots + \frac{x^2}{2} + x + 1$,then $f'(0) = $

Find the derivative: $\frac{d}{dx}(x e^{x^2}) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module