# How do you differentiate f(x)=e^(1+4x)*sin(5-x) using the product rule?

Oct 2, 2016

$f ' \left(x\right) = {e}^{1 + 4 x} \left\{4 \sin \left(5 - x\right) - \cos \left(5 - x\right)\right\}$.

#### Explanation:

Let, f(x)=e^(1+4x)sin(5-x)=uv, say, where, u=e^(1+4x), &, v=sin(5-x).

Using Product Rule, $f ' \left(x\right) = u \frac{\mathrm{dv}}{\mathrm{dx}} + v \frac{\mathrm{du}}{\mathrm{dx}} \ldots \ldots \ldots \ldots \left(\star\right)$

$u = {e}^{1 + 4 x} \Rightarrow \frac{\mathrm{du}}{\mathrm{dx}} = {e}^{1 + 4 x} \cdot \frac{d}{\mathrm{dx}} \left(1 + 4 x\right) \ldots \ldots \ldots . . \text{[Chain Rule]}$

$\therefore \frac{\mathrm{du}}{\mathrm{dx}} = 4 {e}^{1 + 4 x} \ldots \ldots \ldots \ldots \left(1\right)$.

$v = \sin \left(5 - x\right) \Rightarrow \frac{\mathrm{dv}}{\mathrm{dx}} = \cos \left(5 - x\right) \frac{d}{\mathrm{dx}} \left(5 - x\right) \ldots \ldots \text{[Chain Rule]}$

$\therefore \frac{\mathrm{dv}}{\mathrm{dx}} = - \cos \left(5 - x\right) \ldots \ldots \ldots \ldots \ldots \ldots . \left(2\right)$

Using (1) & (2)" in "(star),

$f ' \left(x\right) = 4 {e}^{1 + 4 x} \sin \left(5 - x\right) - \cos \left(5 - x\right) {e}^{1 + 4 x}$.

$\therefore f ' \left(x\right) = {e}^{1 + 4 x} \left\{4 \sin \left(5 - x\right) - \cos \left(5 - x\right)\right\}$.