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] = $

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