The derivative of y=(sin x)^{x^2} with respec | vedclass

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(D) Given $y=(\sin x)^{x^2}$.Taking the natural logarithm on both sides:$\log y = x^2 \log(\sin x)$.Differentiating both sides with respect to $x$ usi

Vedclass · Accepted Answer

$x^2(\sin x)^{x^2-1} \cos x+2 x(\sin x)^{x^2} \log (\sin x)$

Vedclass · Answer

(D) Given $y=(\sin x)^{x^2}$.Taking the natural logarithm on both sides:$\log y = x^2 \log(\sin x)$.Differentiating both sides with respect to $x$ using the product rule:$\frac{1}{y} \frac{dy}{dx} = x^2 \cdot \frac{d}{dx}(\log(\sin x)) + \log(\sin x) \cdot \frac{d}{dx}(x^2)$.$\frac{1}{y} \frac{dy}{dx} = x^2 \cdot \frac{1}{\sin x} \cdot \cos x + \log(\sin x) \cdot 2x$.$\frac{1}{y} \frac{dy}{dx} = x^2 \cot x + 2x \log(\sin x)$.Multiplying by $y$:$\frac{dy}{dx} = y \cdot (x^2 \cot x + 2x \log(\sin x))$.Substituting $y = (\sin x)^{x^2}$:$\frac{dy}{dx} = (\sin x)^{x^2} (x^2 \cot x + 2x \log(\sin x))$.Distributing the term:$\frac{dy}{dx} = x^2 (\sin x)^{x^2} \frac{\cos x}{\sin x} + 2x (\sin x)^{x^2} \log(\sin x)$.$\frac{dy}{dx} = x^2 (\sin x)^{x^2-1} \cos x + 2x (\sin x)^{x^2} \log(\sin x)$.

The derivative of $y=(\sin x)^{x^2}$ with respect to $x$ is

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