If f(x) = frac{1}{x^3} int_5^x (2u^2 - u f'(u | vedclass

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(D) Given, $f(x) = \frac{1}{x^3} \int_5^x (2u^2 - u f'(u)) du$ Multiplying by $x^3$, we get $x^3 f(x) = \int_5^x (2u^2 - u f'(u)) du$ Differentiating

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$\frac{5}{13}$

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(D) Given, $f(x) = \frac{1}{x^3} \int_5^x (2u^2 - u f'(u)) du$ Multiplying by $x^3$, we get $x^3 f(x) = \int_5^x (2u^2 - u f'(u)) du$ Differentiating both sides with respect to $x$ using the Leibniz rule: $x^3 f'(x) + 3x^2 f(x) = 2x^2 - x f'(x)$ Rearranging the terms to solve for $f'(x)$: $x^3 f'(x) + x f'(x) = 2x^2 - 3x^2 f(x)$ $f'(x)(x^3 + x) = 2x^2 - 3x^2 f(x)$ $f'(x) = \frac{2x^2 - 3x^2 f(x)}{x^3 + x}$ At $x = 5$, we know $f(5) = \frac{1}{5^3} \int_5^5 (2u^2 - u f'(u)) du = 0$. Substituting $x = 5$ and $f(5) = 0$ into the expression for $f'(x)$: $f'(5) = \frac{2(5)^2 - 3(5)^2(0)}{5^3 + 5} = \frac{50}{125 + 5} = \frac{50}{130} = \frac{5}{13}$

If $f(x) = \frac{1}{x^3} \int_5^x (2u^2 - u f'(u)) du$, then $f'(5) = $

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