For x

Question

(B) Given $x < 0$,we have $|x| = -x$. Substituting this into the expression,we get $y = (-x)^x$. Taking the natural logarithm on both sides: $\log y =

Vedclass · Accepted Answer

$(-x)^x [1 + \log(-x)]$

Vedclass · Answer

(B) Given $x Substituting this into the expression,we get $y = (-x)^x$. Taking the natural logarithm on both sides: $\log y = x \log(-x)$. Differentiating both sides with respect to $x$ using the product rule: $\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} [x] \cdot \log(-x) + x \cdot \frac{d}{dx} [\log(-x)]$. $\frac{1}{y} \frac{dy}{dx} = 1 \cdot \log(-x) + x \cdot \frac{1}{-x} \cdot (-1) = \log(-x) + 1$. Therefore,$\frac{dy}{dx} = y [\log(-x) + 1] = (-x)^x [1 + \log(-x)]$.

For $x < 0$,$\frac{d}{dx} [|x|^x] = $

Similar Questions

The derivative of $\log |x|$ is

If $y=2^{ax}$ and $\left(\frac{dy}{dx}\right)_{x=1}=\log 256$,then $a=$

If $y = \log_{10} x + \log_{x} 10 + \log_{x} x + \log_{10} 10$,then $\frac{dy}{dx}$ is equal to

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

Find the derivative: $\frac{d}{dx}(\log_7(\log_7 x))$

Vedclass Test Series

Exam Paper Generator

Online Exam Module