If y = (x^2 + 1)^{sin x} for x 0 such that | vedclass

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(D) Given $y = (x^2 + 1)^{\sin x}$.Taking natural logarithm on both sides: $\ln y = \sin x \cdot \ln(x^2 + 1)$.Differentiating with respect to $x$: $\

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strictly decreasing

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(D) Given $y = (x^2 + 1)^{\sin x}$.Taking natural logarithm on both sides: $\ln y = \sin x \cdot \ln(x^2 + 1)$.Differentiating with respect to $x$: $\frac{1}{y} \frac{dy}{dx} = \cos x \cdot \ln(x^2 + 1) + \sin x \cdot \frac{1}{x^2 + 1} \cdot 2x$.Thus, $\frac{dy}{dx} = y [\frac{2x \sin x}{x^2 + 1} + \cos x \cdot \ln(x^2 + 1)]$.Comparing this with the given form $\frac{dy}{dx} = y [\frac{2x \sin x}{g(x)} + \cos x \cdot \log(g(x))]$, we identify $g(x) = x^2 + 1$.We need to analyze the function $f(x) = \frac{1}{g(x)} = \frac{1}{x^2 + 1}$.For $x > 0$, as $x$ increases, $x^2 + 1$ increases.Since $x^2 + 1$ is in the denominator and is always positive, as $x^2 + 1$ increases, $\frac{1}{x^2 + 1}$ decreases.Specifically, $f'(x) = -\frac{2x}{(x^2 + 1)^2}$. Since $x > 0$, $f'(x) Therefore, the function is strictly decreasing.

If $y = (x^2 + 1)^{\sin x}$ for $x > 0$ such that $\frac{dy}{dx} = y [\frac{2x \sin x}{g(x)} + \cos x \cdot \log(g(x))]$, then the function $\frac{1}{g(x)}$ is...

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