# How do you find the derivative of r(x)= (0.3x-4.9x^-1)^0.5?

Dec 26, 2017

$\frac{0.3 + \frac{4.9}{x} ^ 2}{2 \sqrt{0.3 x - \frac{4.9}{x}}}$

#### Explanation:

Let $u = 0.3 x - 4.9 {x}^{-} 1$

$\sqrt{u} = \sqrt{0.3 x - 4.9 {x}^{-} 1}$

Using the chain rule,

Differentiate $\sqrt{u}$ (the function) into 1/(2sqrt(u) and $0.3 x - 4.9 {x}^{-} 1$ (the inside) into $0.3 + \frac{4.9}{x} ^ 2$.

Multiply the two derivatives together to get 1/(2sqrt(u)$\left(0.3 + \frac{4.9}{x} ^ 2\right)$.

Now, substitute $\sqrt{u} = \sqrt{0.3 x - 4.9 {x}^{-} 1}$ for 1/(2sqrt(u).

1/(2sqrt(0.3x-4.9x^-1)$\left(0.3 + \frac{4.9}{x} ^ 2\right)$ which simplifies into:
$\frac{0.3 + \frac{4.9}{x} ^ 2}{2 \sqrt{0.3 x - \frac{4.9}{x}}}$