# How do you differentiate f(x)= (x^3+2x+1)*sqrtsinx using the product rule?

Dec 15, 2015

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

#### Explanation:

The product rule is:

If $f \left(x\right) = g \left(x\right) \cdot h \left(x\right)$, then the derivative of $f \left(x\right)$ is

$f ' \left(x\right) = g ' \left(x\right) \cdot h \left(x\right) + g \left(x\right) \cdot h ' \left(x\right)$

$g \left(x\right) = {x}^{3} + 2 x + 1$

and

$h \left(x\right) = \sqrt{\sin x} = {\left(\sin x\right)}^{\frac{1}{2}}$

So, let's compute the derivatives of $g \left(x\right)$ and $h \left(x\right)$:

$g ' \left(x\right) = 3 {x}^{2} + 2$

To build the derivative of $h \left(x\right)$, you need to apply the chain rule:

$h ' \left(x\right) = \frac{1}{2} {\left(\sin x\right)}^{- \frac{1}{2}} \cdot \cos x = \frac{1}{2 \sqrt{\sin x}} \cdot \cos x$

Now, the only thing left to do is use the product rule:

$f ' \left(x\right) = g ' \left(x\right) \cdot h \left(x\right) + g \left(x\right) \cdot h ' \left(x\right)$

$\textcolor{w h i t e}{\times \xi i} = \left(3 {x}^{2} + 2\right) \sqrt{\sin x} + \left({x}^{3} + 2 x + 1\right) \cdot \frac{1}{2 \sqrt{\sin x}} \cdot \cos x$