If y=x^{log x}+(log x)^x, x1 then left(frac{ | vedclass

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

Question

(D) Let $y = u + v$, where $u = x^{\log x}$ and $v = (\log x)^x$. Taking log on both sides for $u$: $\log u = (\log x)(\log x) = (\log x)^2$. Differen

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(D) Let $y = u + v$, where $u = x^{\log x}$ and $v = (\log x)^x$. Taking log on both sides for $u$: $\log u = (\log x)(\log x) = (\log x)^2$. Differentiating with respect to $x$: $\frac{1}{u} \frac{du}{dx} = 2(\log x) \cdot \frac{1}{x} \implies \frac{du}{dx} = x^{\log x} \cdot \frac{2 \log x}{x}$. At $x=e$: $\frac{du}{dx} = e^{\log e} \cdot \frac{2 \log e}{e} = e^1 \cdot \frac{2}{e} = 2$. Now for $v = (\log x)^x$: $\log v = x \log(\log x)$. Differentiating with respect to $x$: $\frac{1}{v} \frac{dv}{dx} = 1 \cdot \log(\log x) + x \cdot \frac{1}{\log x} \cdot \frac{1}{x} = \log(\log x) + \frac{1}{\log x}$. $\frac{dv}{dx} = (\log x)^x \left[ \log(\log x) + \frac{1}{\log x} \right]$. At $x=e$: $\frac{dv}{dx} = (\log e)^e \left[ \log(\log e) + \frac{1}{\log e} \right] = 1^e [ \log(1) + 1 ] = 1 \cdot [0 + 1] = 1$. Thus, $\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} = 2 + 1 = 3$.

If $y=x^{\log x}+(\log x)^x, x>1$ then $\left(\frac{d y}{d x}\right)_{x=e}=$

Similar Questions

The derivative of $(\log x)^{x}$ with respect to $\log x$ is

If $y = \frac{\sqrt[3]{1 + 3x} \sqrt[4]{1 + 4x} \sqrt[5]{1 + 5x}}{\sqrt[7]{1 + 7x} \sqrt[8]{1 + 8x}}$,then $y'(0)$ is equal to

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

If $y = (x^2 + 1)^{\sin x}$ for $x > 0$ such that $\frac{dy}{dx} = y [\frac{2x \sin x}{g(x)} + \cos x \cdot \log(g(x))]$, then the function $\frac{1}{g(x)}$ is...

Differentiate $\sqrt{\frac{(x-3)(x^{2}+4)}{3x^{2}+4x+5}}$ with respect to $x$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module