If y = f(x)^{g(x)} and frac{dy}{dx} = y[H(x)f | vedclass

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(B) Given $y = f(x)^{g(x)}$. Taking natural logarithm on both sides, we get $\log y = g(x) \log f(x)$.Differentiating both sides with respect to $x$,

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$\frac{[\log f(x)]^2}{2} + c$

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(B) Given $y = f(x)^{g(x)}$. Taking natural logarithm on both sides, we get $\log y = g(x) \log f(x)$.Differentiating both sides with respect to $x$, we have $\frac{1}{y} \frac{dy}{dx} = g'(x) \log f(x) + g(x) \frac{f'(x)}{f(x)}$.Thus, $\frac{dy}{dx} = y \left[ \frac{g(x)}{f(x)} f'(x) + \log f(x) g'(x) \right]$.Comparing this with the given expression $\frac{dy}{dx} = y[H(x)f'(x) + G(x)g'(x)]$, we identify $H(x) = \frac{g(x)}{f(x)}$ and $G(x) = \log f(x)$.Now, we need to evaluate the integral $I = \int \frac{G(x)H(x)f'(x)}{g(x)} dx$.Substituting the identified functions, $I = \int \frac{\log f(x) \cdot \frac{g(x)}{f(x)} \cdot f'(x)}{g(x)} dx = \int \frac{\log f(x) f'(x)}{f(x)} dx$.Let $u = \log f(x)$, then $du = \frac{f'(x)}{f(x)} dx$.Substituting these into the integral, $I = \int u du = \frac{u^2}{2} + c = \frac{[\log f(x)]^2}{2} + c$.

If $y = f(x)^{g(x)}$ and $\frac{dy}{dx} = y[H(x)f'(x) + G(x)g'(x)]$, then $\int \frac{G(x)H(x)f'(x)}{g(x)} dx =$

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