If y=frac{(x+1)^2 sqrt{x-1}}{(x+4)^3 e^x}, th | vedclass

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

Question

(C) Given $y=\frac{(x+1)^2 \sqrt{x-1}}{(x+4)^3 e^x}$.Taking the natural logarithm on both sides:$\log y = \log \left( \frac{(x+1)^2 (x-1)^{1/2}}{(x+4)

Vedclass · Accepted Answer

$\frac{(x+1)^2 \sqrt{x-1}}{(x+4)^3 e^x}\left[\frac{2}{x+1}+\frac{1}{2(x-1)}-\frac{3}{x+4}-1\right]$

Vedclass · Answer

(C) Given $y=\frac{(x+1)^2 \sqrt{x-1}}{(x+4)^3 e^x}$.Taking the natural logarithm on both sides:$\log y = \log \left( \frac{(x+1)^2 (x-1)^{1/2}}{(x+4)^3 e^x} \right)$Using logarithmic properties $\log(ab) = \log a + \log b$ and $\log(a/b) = \log a - \log b$:$\log y = 2 \log(x+1) + \frac{1}{2} \log(x-1) - 3 \log(x+4) - x \log e$Since $\log e = 1$, we have:$\log y = 2 \log(x+1) + \frac{1}{2} \log(x-1) - 3 \log(x+4) - x$Differentiating both sides with respect to $x$:$\frac{1}{y} \frac{dy}{dx} = \frac{2}{x+1} + \frac{1}{2(x-1)} - \frac{3}{x+4} - 1$Therefore, $\frac{dy}{dx} = y \left[ \frac{2}{x+1} + \frac{1}{2(x-1)} - \frac{3}{x+4} - 1 \right]$Substituting the value of $y$:$\frac{dy}{dx} = \frac{(x+1)^2 \sqrt{x-1}}{(x+4)^3 e^x} \left[ \frac{2}{x+1} + \frac{1}{2(x-1)} - \frac{3}{x+4} - 1 \right]$

If $y=\frac{(x+1)^2 \sqrt{x-1}}{(x+4)^3 e^x}$, then $\frac{d y}{d x}=$

Similar Questions

If $y=(1+x)(1+x^2)(1+x^4) \dots (1+x^{2n})$,then the value of $\frac{dy}{dx}$ at $x=0$ is

If $f(x)=x^{\tan x}+(\tan x)^{x}$, then $f^{\prime}\left(\frac{\pi}{4}\right)=$

Differentiate the following with respect to $x$: $\cos x \cdot \cos 2x \cdot \cos 3x$

Differentiate the function $y = x^{x} - 2^{\sin x}$ with respect to $x$.

Assertion $(A)$: $\frac{d}{d x}\left(\frac{x^2 \sin x}{\log x}\right)=\frac{x^2 \sin x}{\log x} \left(\cot x+\frac{2}{x}-\frac{1}{x \log x}\right)$
Reason $(R)$: $\frac{d}{d x}\left(\frac{u v}{w}\right)=\frac{u v}{w}\left[\frac{u^{\prime}}{u}+\frac{v^{\prime}}{v}-\frac{w^{\prime}}{w}\right]$

Vedclass Test Series

Exam Paper Generator

Online Exam Module