# How do you differentiate g(x) = (1/(x^3-1))*sqrt(1+e^(x)) using the product rule?

Jan 13, 2016

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

#### Explanation:

The product rule states: if

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

the derivative of $g \left(x\right)$ can be computed as follows:

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

$f \left(x\right) = \frac{1}{{x}^{3} - 1} \text{ }$ and

$h \left(x\right) = \sqrt{1 + {e}^{x}}$

The first thing you need to do is differentiate $f \left(x\right)$ and $h \left(x\right)$.

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Let's start with $f \left(x\right)$. Here, the chain rule is helpful. You can write use it as follows:

$f \left(x\right) = \frac{1}{u}$ where $u = {x}^{3} - 1$

According to the chain rule, the derivative is the derivative of $\frac{1}{u}$ multiplied with the derivative of $u$.

$\left[\frac{1}{u}\right] ' = \left[{u}^{- 1}\right] ' = - {u}^{- 2} = - \frac{1}{u} ^ 2 = - \frac{1}{{x}^{3} - 1} ^ 2$

$\left[{x}^{3} - 1\right] ' = 3 {x}^{2}$

Thus,

$f ' \left(x\right) = - \frac{1}{{x}^{3} - 1} ^ 2 \cdot 3 {x}^{2} = - \frac{3 {x}^{2}}{{x}^{3} - 1} ^ 2$

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Now, you still need to differentiate $h \left(x\right)$. The chain rule can help here as well:

$h \left(x\right) = \sqrt{v}$ where $v = 1 + {e}^{x}$

The derivative of $h \left(x\right)$ is the derivative of $\sqrt{v}$ multiplied with the derivative of $v$.

$\left[\sqrt{v}\right] ' = \left[{v}^{\frac{1}{2}}\right] ' = \frac{1}{2} {v}^{- \frac{1}{2}} = \frac{1}{2 \sqrt{v}} = \frac{1}{2 \sqrt{1 + {e}^{x}}}$

$\left[1 + {e}^{x}\right] ' = {e}^{x}$

This means that the derivative of $h \left(x\right)$ is:

$h ' \left(x\right) = \frac{1}{2 \sqrt{1 + {e}^{x}}} \cdot {e}^{x} = {e}^{x} / \left(2 \sqrt{1 + {e}^{x}}\right)$

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

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

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

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