How do you integrate f(x)=e^xsinxcosx using the product rule?

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Steve M Share
Jan 11, 2017

$f ' \left(x\right) = {e}^{x} \left(\sin x \cos x + {\cos}^{2} x - {\sin}^{2} x\right)$

Explanation:

If you are studying maths, then you should learn the Product Rule for Differentiation, and practice how to use it:

$\frac{d}{\mathrm{dx}} \left(u v\right) = u \frac{\mathrm{dv}}{\mathrm{dx}} + \frac{\mathrm{du}}{\mathrm{dx}} v$, or, $\left(u v\right) ' = \left(\mathrm{du}\right) v + u \left(\mathrm{dv}\right)$

I was taught to remember the rule in words; "The first times the derivative of the second plus the derivative of the first times the second ".

This can be extended to three products:

$\frac{d}{\mathrm{dx}} \left(u v w\right) = u v \frac{\mathrm{dw}}{\mathrm{dx}} + u \frac{\mathrm{dv}}{\mathrm{dx}} w + \frac{\mathrm{du}}{\mathrm{dx}} v w$

So with $f \left(x\right) = {e}^{x} \sin x \cos x$ we have;

$\left\{\begin{matrix}\text{Let "u = e^x & => & (du)/dx = e^x \\ "And "v=sinx & => & (dv)/dx = cosx \\ "And } w = \cos x & \implies & \frac{\mathrm{dw}}{\mathrm{dx}} = - \sin x\end{matrix}\right.$

Applying the product rule we get:

$\frac{d}{\mathrm{dx}} \left(u v w\right) = \frac{\mathrm{du}}{\mathrm{dx}} v w + u \frac{\mathrm{dv}}{\mathrm{dx}} w + + u v \frac{\mathrm{dw}}{\mathrm{dx}}$

$\therefore f ' \left(x\right) = \left({e}^{x}\right) \left(\sin x\right) \left(\cos x\right) + \left({e}^{x}\right) \left(\cos x\right) \left(\cos x\right) + \left({e}^{x}\right) \left(\sin x\right) \left(- \sin x\right)$
$\text{ } = {e}^{x} \left(\sin x \cos x + {\cos}^{2} x - {\sin}^{2} x\right)$

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