# What is the derivative of g(x)=x+(4/x)?

Jul 30, 2018

$g ' \left(x\right) = 1 - \frac{4}{{x}^{2}}$

#### Explanation:

To find the derivative of $g \left(x\right)$, you must differentiate each term in the sum

$g ' \left(x\right) = \frac{d}{\mathrm{dx}} \left(x\right) + \frac{d}{\mathrm{dx}} \left(\frac{4}{x}\right)$

It is easier to see the Power Rule on the second term by rewriting it as

$g ' \left(x\right) = \frac{d}{\mathrm{dx}} \left(x\right) + \frac{d}{\mathrm{dx}} \left(4 {x}^{-} 1\right)$

$g ' \left(x\right) = 1 + 4 \frac{d}{\mathrm{dx}} \left({x}^{-} 1\right)$

$g ' \left(x\right) = 1 + 4 \left(- 1 {x}^{- 1 - 1}\right)$

$g ' \left(x\right) = 1 + 4 \left(- {x}^{- 2}\right)$

$g ' \left(x\right) = 1 - 4 {x}^{-} 2$

Finally, you can rewrite this new second term as a fraction:

$g ' \left(x\right) = 1 - \frac{4}{{x}^{2}}$

Jul 30, 2018

$g ' \left(x\right) = 1 - \frac{4}{{x}^{2}}$

#### Explanation:

What might be daunting is the $\frac{4}{x}$. Luckily, we can rewrite this as $4 {x}^{-} 1$. Now, we have the following:

$\frac{d}{\mathrm{dx}} \left(x + 4 {x}^{-} 1\right)$

We can use the Power Rule here. The exponent comes out front, and the power gets decremented by one. We now have

$g ' \left(x\right) = 1 - 4 {x}^{-} 2$, which can be rewritten as

$g ' \left(x\right) = 1 - \frac{4}{{x}^{2}}$

Hope this helps!