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

Nov 7, 2015

Treat two of the multiplicands as a single function, then apply the product rule twice.

#### Explanation:

The product rule states that $\frac{d}{\mathrm{dx}} f \left(x\right) g \left(x\right) = f ' \left(x\right) g \left(x\right) + f \left(x\right) g ' \left(x\right)$

To apply it in this case, let's let
${f}_{1} \left(x\right) = {e}^{x} + 1$
${f}_{2} \left(x\right) = x + 1$
${f}_{3} \left(x\right) = \sin \left(x\right)$
${f}_{4} \left(x\right) = {f}_{2} \left(x\right) {f}_{3} \left(x\right) = \left(x + 1\right) \sin \left(x\right)$

Now we can write $f \left(x\right) = {f}_{1} \left(x\right) {f}_{4} \left(x\right)$
So, applying the product rule,
$f ' \left(x\right) = {f}_{1} ' \left(x\right) {f}_{4} \left(x\right) + {f}_{1} \left(x\right) {f}_{4} ' \left(x\right)$

But ${f}_{4} \left(x\right) = {f}_{2} \left(x\right) {f}_{3} \left(x\right)$, so once again we apply the product rule
${f}_{4} ' \left(x\right) = {f}_{2} ' \left(x\right) {f}_{3} \left(x\right) + {f}_{2} \left(x\right) {f}_{3} ' \left(x\right)$

Then we can substitute to put $f ' \left(x\right)$ in terms of ${f}_{1} \left(x\right)$, ${f}_{2} \left(x\right)$, and ${f}_{3} \left(x\right)$ and their derivatives.

$f ' \left(x\right) = {f}_{1} ' \left(x\right) \left({f}_{2} \left(x\right) {f}_{3} \left(x\right)\right) + {f}_{1} \left(x\right) \left({f}_{2} ' \left(x\right) {f}_{3} \left(x\right) + {f}_{2} \left(x\right) {f}_{3} ' \left(x\right)\right)$

Now we calculate ${f}_{1} ' \left(x\right)$, ${f}_{2} ' \left(x\right)$, and ${f}_{3} ' \left(x\right)$.
${f}_{1} ' \left(x\right) = {e}^{x}$
${f}_{2} ' \left(x\right) = 1$
${f}_{3} ' \left(x\right) = \cos \left(x\right)$

Substituting this all in, we have
$f ' \left(x\right) = {e}^{x} \left(x + 1\right) \sin \left(x\right) + \left({e}^{x} + 1\right) \left(1 \cdot \sin \left(x\right) + \left(x + 1\right) \cos \left(x\right)\right)$