If y = frac{sqrt{x}(2x + 3)^2}{sqrt{x + 1}},t | vedclass

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

Question

(A) Given $y = \frac{\sqrt{x}(2x + 3)^2}{\sqrt{x + 1}}$.Taking natural logarithm on both sides:$\ln y = \ln \left( \frac{\sqrt{x}(2x + 3)^2}{\sqrt{x +

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given $y = \frac{\sqrt{x}(2x + 3)^2}{\sqrt{x + 1}}$.Taking natural logarithm on both sides:$\ln y = \ln \left( \frac{\sqrt{x}(2x + 3)^2}{\sqrt{x + 1}} \right)$$\ln y = \ln(\sqrt{x}) + \ln((2x + 3)^2) - \ln(\sqrt{x + 1})$$\ln y = \frac{1}{2} \ln x + 2 \ln(2x + 3) - \frac{1}{2} \ln(x + 1)$Differentiating both sides with respect to $x$:$\frac{1}{y} \frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{x} + 2 \cdot \frac{1}{2x + 3} \cdot 2 - \frac{1}{2} \cdot \frac{1}{x + 1}$$\frac{1}{y} \frac{dy}{dx} = \frac{1}{2x} + \frac{4}{2x + 3} - \frac{1}{2(x + 1)}$Therefore,$\frac{dy}{dx} = y \left[ \frac{1}{2x} + \frac{4}{2x + 3} - \frac{1}{2(x + 1)} \right]$.

If $y = \frac{\sqrt{x}(2x + 3)^2}{\sqrt{x + 1}}$,then $\frac{dy}{dx} = $

Similar Questions

If $y(\cos x)^{\sin x}=(\sin x)^{\sin x}$, then the value of $\frac{dy}{dx}$ at $x=\frac{\pi}{4}$ is

If $f(x)=\frac{\cos ^2 x}{1+\sin ^2 x}$, then $f\left(\frac{\pi}{4}\right)-3 f^{\prime}\left(\frac{\pi}{4}\right)=$

Differentiate the function with respect to $x$: $(5x)^{3 \cos 2x}$

If $y=\sqrt{e^{\sqrt{x}}}$,then $\frac{d y}{d x}=$

Differentiate the function with respect to $x$: $(\log x)^{\cos x}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module