How do you differentiate f(x)= (cos x)*e^-x*sqrt(3x) using the product rule?

Nov 10, 2015

$\frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{- x} \setminus \cdot \sqrt{3 x} \left(\cos \left(x\right) \cdot \frac{1 - 2 x}{2 x} - \sin \left(x\right)\right)$

Explanation:

So we have

$y = \cos \left(x\right) \cdot {e}^{- x} \cdot \sqrt{3 x}$

To simplify things, let's say ${e}^{- x} \sqrt{3 x} = u$

So we have

$y = u \cdot \cos \left(x\right)$

Using the product rule we have

$\frac{\mathrm{dy}}{\mathrm{dx}} = u \frac{d}{\mathrm{dx}} \left(\cos \left(x\right)\right) + \cos \left(x\right) \frac{\mathrm{du}}{\mathrm{dx}}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = - u \cdot \sin \left(x\right) + \cos \left(x\right) \frac{\mathrm{du}}{\mathrm{dx}}$

And now we differentiate $u$

$u = {e}^{- x} \cdot \sqrt{3 x} = {e}^{- x} \cdot \left(\sqrt{3} {x}^{\frac{1}{2}}\right)$

$\frac{\mathrm{du}}{\mathrm{dx}} = {e}^{- x} \frac{d}{\mathrm{dx}} \left(\sqrt{3} {x}^{\frac{1}{2}}\right) + \sqrt{3 x} \frac{d}{\mathrm{dx}} \left({e}^{- x}\right)$

$\frac{\mathrm{du}}{\mathrm{dx}} = {e}^{- x} \cdot \frac{\sqrt{3}}{2} {x}^{- \frac{1}{2}} - \sqrt{3 x} \cdot {e}^{- x}$
$\frac{\mathrm{du}}{\mathrm{dx}} = {e}^{- x} \left(\frac{\sqrt{3}}{2 \sqrt{x}} - \sqrt{3 x}\right)$

Using some algebra to change that we have

$\frac{\mathrm{du}}{\mathrm{dx}} = {e}^{- x} \left(\frac{\sqrt{3 x}}{2 x} - \sqrt{3 x}\right)$
$\frac{\mathrm{du}}{\mathrm{dx}} = {e}^{- x} \cdot \sqrt{3 x} \left(\frac{1}{2 x} - 1\right)$
$\frac{\mathrm{du}}{\mathrm{dx}} = \frac{u \left(1 - 2 x\right)}{2 x}$

Substitute that back in the original derivative

$\frac{\mathrm{dy}}{\mathrm{dx}} = - u \cdot \sin \left(x\right) + \cos \left(x\right) \cdot u \setminus \cdot \frac{1 - 2 x}{2 x}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = u \left(\cos \left(x\right) \cdot \frac{1 - 2 x}{2 x} - \sin \left(x\right)\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{- x} \setminus \cdot \sqrt{3 x} \left(\cos \left(x\right) \cdot \frac{1 - 2 x}{2 x} - \sin \left(x\right)\right)$