# How do you differentiate f(x)=4x*e^x*sinx using the product rule?

Mar 19, 2016

$\frac{d}{\mathrm{dx}} f \left(x\right) = 4 \cdot {e}^{x} \left(\sin x + x \sin x + x \cos x\right)$

#### Explanation:

Product rule states that

$\frac{d}{\mathrm{dx}} \left(f \left(x\right) g \left(x\right) h \left(x\right)\right)$
= $\frac{\mathrm{df} \left(x\right)}{\mathrm{dx}} \cdot g \left(x\right) h \left(x\right) + \frac{\mathrm{dg} \left(x\right)}{\mathrm{dx}} \cdot f \left(x\right) h \left(x\right) + \frac{\mathrm{dh} \left(x\right)}{\mathrm{dx}} \cdot f \left(x\right) g \left(x\right)$

Hence, $\frac{d}{\mathrm{dx}} \left(4 x \cdot {e}^{x} \cdot \sin x\right)$

= $\frac{d \left(4 x\right)}{\mathrm{dx}} \cdot {e}^{x} \cdot \sin x + \frac{{\mathrm{de}}^{x}}{\mathrm{dx}} \cdot 4 x \cdot \sin x + \frac{\mathrm{ds} \in x}{\mathrm{dx}} \cdot 4 x \cdot {e}^{x}$

= $4 \cdot {e}^{x} \cdot \sin x + {e}^{x} \cdot 4 x \cdot \sin x + \cos x \cdot 4 x \cdot {e}^{x}$

= $4 \cdot {e}^{x} \left(\sin x + x \sin x + x \cos x\right)$