If y = x^{x^2},then frac{dy}{dx} = | vedclass

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(B) Given $y = x^{x^2}$.Taking natural logarithm on both sides,we get $\ln y = \ln(x^{x^2})$.Using the property $\ln(a^b) = b \ln a$,we have $\ln y =

Vedclass · Accepted Answer

$x^{x^2} \cdot x \cdot (2 \ln x + 1)$

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(B) Given $y = x^{x^2}$.Taking natural logarithm on both sides,we get $\ln y = \ln(x^{x^2})$.Using the property $\ln(a^b) = b \ln a$,we have $\ln y = x^2 \ln x$.Differentiating both sides with respect to $x$ using the product rule:$\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(x^2) \cdot \ln x + x^2 \cdot \frac{d}{dx}(\ln x)$.$\frac{1}{y} \frac{dy}{dx} = 2x \ln x + x^2 \cdot \frac{1}{x}$.$\frac{1}{y} \frac{dy}{dx} = 2x \ln x + x$.$\frac{dy}{dx} = y(2x \ln x + x)$.Substituting $y = x^{x^2}$ and factoring $x$:$\frac{dy}{dx} = x^{x^2} \cdot x(2 \ln x + 1)$.

If $y = x^{x^2}$,then $\frac{dy}{dx} = $

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