How do you differentiate f(x)=(sinx+lnx)(x^2-3e^x) using the product rule?

Jun 20, 2017

$f ' \left(x\right) = \left(\sin x + \ln x\right) \left(2 x - 3 {e}^{x}\right) + \left(\cos x + \frac{1}{x}\right) \left({x}^{2} - 3 {e}^{x}\right)$

Explanation:

We have:

$f \left(x\right) = \left(\sin x + \ln x\right) \left({x}^{2} - 3 {e}^{x}\right)$

We will apply the Product Rule for Differentiation:

$\frac{d}{\mathrm{dx}} \left(u v\right) = u \frac{\mathrm{dv}}{\mathrm{dx}} + \frac{\mathrm{du}}{\mathrm{dx}} v$, 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 $f \left(x\right) = \left(\sin x + \ln x\right) \left({x}^{2} - 3 {e}^{x}\right)$;

$\left\{\begin{matrix}\text{Let" & u = sinx+lnx & => & (du)/dx = cosx+1/x \\ "And} & v = {x}^{2} - 3 {e}^{x} & \implies & \frac{\mathrm{dv}}{\mathrm{dx}} = 2 x - 3 {e}^{x}\end{matrix}\right.$

Then $\frac{d}{\mathrm{dx}} \left(u v\right) = u \frac{\mathrm{dv}}{\mathrm{dx}} + \frac{\mathrm{du}}{\mathrm{dx}} v$ gives us:

$f ' \left(x\right) = \left(\sin x + \ln x\right) \left(2 x - 3 {e}^{x}\right) + \left(\cos x + \frac{1}{x}\right) \left({x}^{2} - 3 {e}^{x}\right)$