# How do you differentiate  f(x)=x*(9 + 2 x)*sqrt(x^2-3) using the product rule?

Oct 27, 2015

$f ' \left(x\right) = \frac{3 \left(2 {x}^{3} + 6 {x}^{2} - 4 x - 9\right)}{\sqrt{{x}^{2} - 3}}$

#### Explanation:

We have a few options. We can leave the function as it is and use the product rule twice, or you can simplify by multiplying the $x$ through $9 + 2 x$ and using the product rule once. Since the second is easier, lets go with that one.

The product rule tells us that if we have a function $f \left(x\right)$, where;

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

Then;

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

If we simplify the function as mentioned above, by multiplying the $x$ through, we get;

$f \left(x\right) = \left(9 x + 2 {x}^{2}\right) \sqrt{{x}^{2} - 3}$

Now we can split our function into;

$g \left(x\right) = 9 x + 2 {x}^{2}$
$h \left(x\right) = \sqrt{{x}^{2} - 3}$

If we use the power rule to find $g ' \left(x\right)$ we get;

$g ' \left(x\right) = 9 + 4 x$

Finding $h ' \left(x\right)$ is a little bit trickier since we have to use the chain rule . If we have a function $f \left(x\right)$ where;

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

Then;

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

We can rewrite $h \left(x\right)$ to make things a little clearer.

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

Let $u \left(x\right) = {x}^{2} - 3$.

$h \left(u\right) = {u}^{\frac{1}{2}}$

Applying the power rule;

$h ' \left(u\right) = \frac{1}{2} {u}^{\left(\frac{1}{2} - 1\right)} u ' = \frac{1}{\cancel{2}} {\left({x}^{2} - 3\right)}^{- \frac{1}{2}} \left(\cancel{2} x\right) = \frac{x}{\sqrt{{x}^{2} - 3}}$

Now that we have $h '$ and $g '$ we can put everything together.

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

$= \left(9 x + 2 {x}^{2}\right) \frac{x}{\sqrt{{x}^{2} - 3}} + \left(9 + 4 x\right) \sqrt{{x}^{2} - 3}$

All that's left is to simplify.

$= \frac{9 {x}^{2} + 2 {x}^{3}}{\sqrt{{x}^{2} - 3}} + \left(9 + 4 x\right) \sqrt{{x}^{2} - 3}$

$= \frac{9 {x}^{2} + 2 {x}^{3}}{\sqrt{{x}^{2} - 3}} + \left(9 + 4 x\right) \sqrt{{x}^{2} - 3} \cdot \frac{\sqrt{{x}^{2} - 3}}{\sqrt{{x}^{2} - 3}}$

$= \frac{9 {x}^{2} + 2 {x}^{3}}{\sqrt{{x}^{2} - 3}} + \frac{\left(9 + 4 x\right) \left({x}^{2} - 3\right)}{\sqrt{{x}^{2} - 3}}$

$= \frac{\left(9 {x}^{2} + 2 {x}^{3}\right) + 9 {x}^{2} + 4 {x}^{3} - 27 - 12 x}{\sqrt{{x}^{2} - 3}}$

$= \frac{6 {x}^{3} + 18 {x}^{2} - 12 x - 27}{\sqrt{{x}^{2} - 3}}$

$= \frac{3 \left(2 {x}^{3} + 6 {x}^{2} - 4 x - 9\right)}{\sqrt{{x}^{2} - 3}}$

*Note If you choose to use the product rule twice;

$g \left(x\right) = x \left(9 + 2 x\right) = j \left(x\right) \cdot k \left(x\right)$

Where;

$j \left(x\right) = x$
$k \left(x\right) = 9 + 2 x$

So;

$g ' \left(x\right) = j \left(x\right) k ' \left(x\right) + j ' \left(x\right) k \left(x\right)$

And;

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

You end up with the same answer, it just takes more work. This method would be good if you couldn't easily simplify the original function.