Find the derivative of the function: frac{d}{ | vedclass

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

Question

(B) Let $y = (\sin x)^{\log x}$.Taking the natural logarithm on both sides:$\log y = \log x \cdot \log \sin x$.Differentiating both sides with respect

Vedclass · Accepted Answer

$(\sin x)^{\log x} \left[ \frac{1}{x} \log \sin x + \cot x \log x \right]$

Vedclass · Answer

(B) Let $y = (\sin x)^{\log x}$.Taking the natural logarithm on both sides:$\log y = \log x \cdot \log \sin x$.Differentiating both sides with respect to $x$ using the product rule:$\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(\log x) \cdot \log \sin x + \log x \cdot \frac{d}{dx}(\log \sin x)$.$\frac{1}{y} \frac{dy}{dx} = \frac{1}{x} \log \sin x + \log x \cdot \frac{1}{\sin x} \cdot \cos x$.$\frac{1}{y} \frac{dy}{dx} = \frac{1}{x} \log \sin x + \cot x \log x$.Therefore,$\frac{dy}{dx} = y \left[ \frac{1}{x} \log \sin x + \cot x \log x \right]$.Substituting $y = (\sin x)^{\log x}$:$\frac{dy}{dx} = (\sin x)^{\log x} \left[ \frac{1}{x} \log \sin x + \cot x \log x \right]$.

Find the derivative of the function: $\frac{d}{dx} \{(\sin x)^{\log x}\}$

Similar Questions

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

If $y=x^{\sin x}+(\sin x)^x$,then $\frac{d y}{d x}$ at $x=\frac{\pi}{2}$ is

If $x^{p} y^{q}=(x+y)^{p+q}$,then $\frac{d y}{d x}$ is equal to

If ${x^m}{y^n} = 2{(x + y)^{m + n}},$ the value of $\frac{dy}{dx}$ is

If $f(x)=\frac{(x+1) \sinh x}{e^{2 x} \tan x}$ and $\frac{f^{\prime}(x)}{f(x)}=\frac{1}{x+1}+\operatorname{coth} x+g(x)$, then $g(x)=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module