The derivative of (log x)^{sin x} with respec | vedclass

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

Question

(C) Let $f(x) = (\log x)^{\sin x}$ and $g(x) = \cos x$. We need to find $\frac{df}{dg} = \frac{f'(x)}{g'(x)}$. Taking the natural logarithm of $f(x)$:

Vedclass · Accepted Answer

$\frac{-2}{\pi}$

Vedclass · Answer

(C) Let $f(x) = (\log x)^{\sin x}$ and $g(x) = \cos x$. We need to find $\frac{df}{dg} = \frac{f'(x)}{g'(x)}$. Taking the natural logarithm of $f(x)$: $\log(f(x)) = \sin x \cdot \log(\log x)$. Differentiating with respect to $x$: $\frac{1}{f(x)} f'(x) = \cos x \cdot \log(\log x) + \sin x \cdot \frac{1}{\log x} \cdot \frac{1}{x}$. Thus, $f'(x) = (\log x)^{\sin x} \left[ \cos x \cdot \log(\log x) + \frac{\sin x}{x \log x} \right]$. Also, $g'(x) = -\sin x$. Therefore, $\frac{df}{dg} = \frac{(\log x)^{\sin x} [ \cos x \cdot \log(\log x) + \frac{\sin x}{x \log x} ]}{-\sin x}$. At $x = \frac{\pi}{2}$, $\sin x = 1$, $\cos x = 0$, and $\log x = \log(\frac{\pi}{2})$. Substituting these values: $\frac{df}{dg} = \frac{(\log(\frac{\pi}{2}))^1 [ 0 \cdot \log(\log(\frac{\pi}{2})) + \frac{1}{(\frac{\pi}{2}) \log(\frac{\pi}{2})} ]}{-1}$. $= - \frac{1}{\frac{\pi}{2}} = -\frac{2}{\pi}$.

The derivative of $(\log x)^{\sin x}$ with respect to $\cos x$ at $x=\frac{\pi}{2}$ is

Similar Questions

If $y = \log \left[a^{3x} \left(\frac{5-x}{x+4}\right)^{\frac{3}{4}}\right]$,then $\frac{dy}{dx} = $

If the angle between the curves $y = a^x$ and $y = b^x$ is $\alpha$,find the modulus of $\tan \alpha$.

$\frac{d}{dx} \left( \log \sqrt{\frac{1+\sin x}{1-\sin x}} \right) = $

If $f(x)=\log _{x^{2}}\left(\log _{e} x\right)$,then $f^{\prime}(x)$ at $x=e$ is

Differentiate the following with respect to $x$: $\log (\log x)$,where $x > 1$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module