$f (x) = x {sin}^{2} x - x^{3}$ find $f^{'} (x)$

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Solution

$f (x) = x {sin}^{2} x - x^{3} f^{'} (x) = \frac{d}{d x} (x {sin}^{2} x - x^{3}) f^{'} (x) = {sin}^{2} x + 2 x sin x cos x - 3 x^{2}$

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f (x) + f^{'} (x) + 2

f (x) = x sin 2 x

f (x) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 2 {cos}^{2} x & sin 2 x & - sin x sin 2 x & 2 {sin}^{2} x & cos x sin x & - cos x & 0 \end{matrix} ∣ ∣ ∣ ∣ ∣

\int_{0}^{π / 2} [f (x) + f^{'} (x)] d x

f (x) = x^{3} - 27

f^{'} (x)

f (x) = \frac{1}{x^{3}}

f^{^{'}} (x) =

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f (0) = 0

lim n \to \infty n \sum k = 1 f (k π) = \frac{- P π e^{π}}{(e^{π} - 1)^{2}}

\frac{9}{4} P

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