If y = x^{ln x},then dy/dx equals :- | vedclass

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(A) Given $y = x^{\ln x}$.Taking natural logarithm on both sides,we get $\ln y = \ln(x^{\ln x})$.Using the property $\ln(a^b) = b \ln a$,we have $\ln

Vedclass · Accepted Answer

$2 \ln x \cdot x^{\ln x - 1}$

Vedclass · Answer

(A) Given $y = x^{\ln x}$.Taking natural logarithm on both sides,we get $\ln y = \ln(x^{\ln x})$.Using the property $\ln(a^b) = b \ln a$,we have $\ln y = \ln x \cdot \ln x = (\ln x)^2$.Differentiating both sides with respect to $x$:$\frac{1}{y} \cdot \frac{dy}{dx} = \frac{d}{dx}[(\ln x)^2]$.Using the chain rule,$\frac{d}{dx}[(\ln x)^2] = 2 \ln x \cdot \frac{d}{dx}(\ln x) = 2 \ln x \cdot \frac{1}{x}$.Therefore,$\frac{1}{y} \cdot \frac{dy}{dx} = \frac{2 \ln x}{x}$.Multiplying by $y$,we get $\frac{dy}{dx} = y \cdot \frac{2 \ln x}{x}$.Substituting $y = x^{\ln x}$,we get $\frac{dy}{dx} = x^{\ln x} \cdot \frac{2 \ln x}{x} = 2 \ln x \cdot x^{\ln x - 1}$.

If $y = x^{\ln x}$,then $dy/dx$ equals :-

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