Given \( f(x) = (2 – x)^{-2} \):

Using the chain rule, let \( u = 2 – x \), so \( f(x) = u^{-2} \), and \( \frac{du}{dx} = -1 \).

Then, \( \frac{df}{du} = -2 u^{-3} \), so:

\[ f'(x) = \frac{df}{du} \cdot \frac{du}{dx} = (-2 (2 – x)^{-3}) \cdot (-1) = \frac{2}{(2 – x)^3} \]

Given \( g(x) = x^2 \), so \( g'(x) = 2x \):

Using \( f'(x) = \frac{2}{(2 – x)^3} \):

\[ f'(x)g'(x) = \frac{2}{(2 – x)^3} \cdot 2x = \frac{4x}{(2 – x)^3} \]

Using the product rule: \( (f(x)g(x))’ = f(x)g'(x) + g(x)f'(x) \).

Compute each term:

\[ f(x)g'(x) = \frac{1}{(2 – x)^2} \cdot 2x = \frac{2x}{(2 – x)^2} \]

\[ g(x)f'(x) = x^2 \cdot \frac{2}{(2 – x)^3} = \frac{2x^2}{(2 – x)^3} \]

Combine with common denominator \( (2 – x)^3 \):

\[ f(x)g'(x) + g(x)f'(x) = \frac{2x}{(2 – x)^2} \cdot \frac{2 – x}{2 – x} + \frac{2x^2}{(2 – x)^3} = \frac{2x(2 – x) + 2x^2}{(2 – x)^3} \]

Simplify the numerator:

\[ 2x(2 – x) + 2x^2 = 4x – 2x^2 + 2x^2 = 4x \]

Thus:

\[ \frac{4x}{(2 – x)^3} \]

Rearrange the given equation: \( f(x)g'(x) + g(x)f'(x) = f'(x)g'(x) \).

Move all terms to one side:

\[ f(x)g'(x) + g(x)f'(x) – f'(x)g'(x) = 0 \]

Factorize:

\[ f'(x)(g(x) – g'(x)) + f(x)g'(x) = 0 \]

Rearrange:

\[ f'(x)(g'(x) – g(x)) = f(x)g'(x) \]

Divide by \( f(x)(g'(x) – g(x)) \), since \( f(x) > 0 \) and \( g'(x) \neq g(x) \):

\[ \frac{f'(x)}{f(x)} = \frac{g'(x)}{g'(x) – g(x)} \]

Integrate both sides of: \( \frac{f'(x)}{f(x)} = \frac{g'(x)}{g'(x) – g(x)} \).

Left side:

\[ \int \frac{f'(x)}{f(x)} \, dx = \ln f(x) + C_1 \]

Right side:

\[ \ln f(x) + C_1 = \int \frac{g'(x)}{g'(x) – g(x)} \, dx + C_2 \]

Combine constants: \( C = C_2 – C_1 \).

Thus:

\[ \ln f(x) = \int \frac{g'(x)}{g'(x) – g(x)} \, dx + C \]

Exponentiate:

\[ f(x) = e^{\int \frac{g'(x)}{g'(x) – g(x)} \, dx + C} = e^C e^{\int \frac{g'(x)}{g'(x) – g(x)} \, dx} \]

Let \( A = e^C \), where \( A > 0 \):

\[ f(x) = A e^{\int \frac{g'(x)}{g'(x) – g(x)} \, dx} \]

Given \( g(x) = x e^x \), with \( A = 1 \):

Compute \( g'(x) \):

\[ g'(x) = e^x + x e^x = e^x (1 + x) \]

Compute the integrand:

\[ g'(x) – g(x) = e^x (1 + x) – x e^x = e^x \]

\[ \frac{g'(x)}{g'(x) – g(x)} = \frac{e^x (1 + x)}{e^x} = 1 + x \]

Integrate:

\[ \int (1 + x) \, dx = x + \frac{x^2}{2} + C \]

Thus:

\[ f(x) = e^{x + \frac{x^2}{2} + C} = e^C e^{\frac{x^2}{2} + x} \]

Since \( A = 1 \), let \( e^C = 1 \), so:

\[ f(x) = e^{\frac{x^2}{2} + x} \]

Given \( g(x) = \sin x + \cos x \), over \( 0 < x < \pi \), with \( A = 1 \):

Compute \( g'(x) \):

\[ g'(x) = \cos x – \sin x \]

Compute the integrand:

\[ g'(x) – g(x) = (\cos x – \sin x) – (\sin x + \cos x) = -2 \sin x \]

\[ \frac{g'(x)}{g'(x) – g(x)} = \frac{\cos x – \sin x}{-2 \sin x} = \frac{\sin x – \cos x}{2 \sin x} = \frac{1}{2} \left( \frac{\sin x}{\sin x} – \frac{\cos x}{\sin x} \right) = \frac{1}{2} – \frac{1}{2} \cot x \]

Integrate:

\[ \int \left( \frac{1}{2} – \frac{1}{2} \cot x \right) \, dx = \frac{1}{2} x – \frac{1}{2} \int \cot x \, dx \]

Since \( \int \cot x \, dx = \ln |\sin x| + C_1 \), and \( \sin x > 0 \) for \( 0 < x < \pi \):

\[ \int \left( \frac{1}{2} – \frac{1}{2} \cot x \right) \, dx = \frac{1}{2} x – \frac{1}{2} \ln \sin x + C \]

Thus:

\[ f(x) = e^{\frac{1}{2} x – \frac{1}{2} \ln \sin x + C} = e^C e^{\frac{x – \ln \sin x}{2}} = e^C \sqrt{\frac{e^x}{\sin x}} \]

Since \( A = 1 \), let \( e^C = 1 \), so:

\[ f(x) = \sqrt{\frac{e^x}{\sin x}} \]

Express in the form \( f(x) = \sqrt{e^{h(x)}} \):

\[ \sqrt{\frac{e^x}{\sin x}} = \sqrt{e^x \cdot e^{-\ln \sin x}} = \sqrt{e^{x – \ln \sin x}} \]

Thus, \( h(x) = x – \ln \sin x \).

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