If y = e^{4x} left( frac{x-4}{x+3} right)^{fr | vedclass

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

Question

(A) Given $y = e^{4x} \left( \frac{x-4}{x+3} \right)^{\frac{3}{4}}$.Taking natural logarithm on both sides:$\ln y = \ln \left( e^{4x} \left( \frac{x-4

Vedclass · Accepted Answer

$\frac{dy}{dx} = y \left[ 4 + \frac{21}{4(x-4)(x+3)} \right]$

Vedclass · Answer

(A) Given $y = e^{4x} \left( \frac{x-4}{x+3} \right)^{\frac{3}{4}}$.Taking natural logarithm on both sides:$\ln y = \ln \left( e^{4x} \left( \frac{x-4}{x+3} \right)^{\frac{3}{4}} \right)$$\ln y = \ln(e^{4x}) + \ln \left( \frac{x-4}{x+3} \right)^{\frac{3}{4}}$$\ln y = 4x + \frac{3}{4} [\ln(x-4) - \ln(x+3)]$Differentiating with respect to $x$:$\frac{1}{y} \frac{dy}{dx} = 4 + \frac{3}{4} \left( \frac{1}{x-4} - \frac{1}{x+3} \right)$$\frac{1}{y} \frac{dy}{dx} = 4 + \frac{3}{4} \left( \frac{(x+3) - (x-4)}{(x-4)(x+3)} \right)$$\frac{1}{y} \frac{dy}{dx} = 4 + \frac{3}{4} \left( \frac{7}{(x-4)(x+3)} \right)$$\frac{1}{y} \frac{dy}{dx} = 4 + \frac{21}{4(x-4)(x+3)}$Therefore,$\frac{dy}{dx} = y \left[ 4 + \frac{21}{4(x-4)(x+3)} \right]$.

If $y = e^{4x} \left( \frac{x-4}{x+3} \right)^{\frac{3}{4}}$,then $\frac{dy}{dx} = $

Similar Questions

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 $(R)$: $\frac{d}{d x}\left(\frac{u v}{w}\right)=\frac{u v}{w}\left[\frac{u^{\prime}}{u}+\frac{v^{\prime}}{v}-\frac{w^{\prime}}{w}\right]$

If $y = f(x)^{g(x)}$ and $\frac{dy}{dx} = y[H(x)f'(x) + G(x)g'(x)]$, then $\int \frac{G(x)H(x)f'(x)}{g(x)} dx =$

If $y = \sqrt{\frac{x^4 \sqrt{3x-5}}{(x^2-3)(2x-3)}}$,then $\left(\frac{dy}{dx}\right)_{x=2} = $

If $f(x) = (|x|)^{|\sin x|}$,then $f'\left( -\frac{\pi}{4} \right) = $

If $y(x) = x^x, x > 0$,then $y^{\prime \prime}(2) - 2y^{\prime}(2)$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module