# Math Help - Calculating second derivative

1. ## Calculating second derivative

I have a function f(x) = sqrt(0.5*x^2 + 4) - 2

I calculate the first derivative and get

f'(x) = x / sqrt(0.5x^2 + 4)

But I have a hard time trying to figure out the second derivative f''(x).

How do I proceed?

2. Originally Posted by Tikitac
I have a function f(x) = sqrt(0.5*x^2 + 4) - 2

I calculate the first derivative and get

f'(x) = 2x / sqrt(0.5x^2 + 4)

But I have a hard time trying to figure out the second derivative f''(x).

How do I proceed?
Well, your first derivative isn't right anyway.

Here's a helpful hint: Use fractions, it's a little clearer that way.
$f(x) = \sqrt{\frac{1}{2}x^2 + 4} - 2$

So
$f^{\prime}(x) = \frac{1}{2} \cdot \frac{1}{\sqrt{\frac{1}{2}x^2 + 4}} \cdot \left ( \frac{1}{2} \cdot 2x \right )$

Now to "pretty" this up a bit:
$f^{\prime}(x) = \frac{x}{2\sqrt{\frac{1}{2}x^2 + 4}}$

$f^{\prime}(x) = \frac{x}{\sqrt{4 \cdot \frac{1}{2}x^2 + 4 \cdot 4}}$

$f^{\prime}(x) = \frac{x}{\sqrt{2x^2 + 16}}$

For your second derivative, use the quotient rule:
$f^{\prime \prime}(x) = \frac{1 \cdot \sqrt{2x^2 + 16} - x \cdot \left ( \frac{1}{2} \cdot \frac{1}{\sqrt{2x^2 + 16}} \cdot (2 \cdot 2x) \right )}{2x^2 + 16}$

$f^{\prime \prime}(x) = \frac{\sqrt{2x^2 + 16} - \frac{2x^2}{\sqrt{2x^2 + 16}}}{2x^2 + 16}$

$f^{\prime \prime}(x) = \frac{\sqrt{2x^2 + 16} - \frac{2x^2}{\sqrt{2x^2 + 16}}}{2x^2 + 16} \cdot \frac{\sqrt{2x^2 + 16}}{\sqrt{2x^2 + 16}}$

$f^{\prime \prime}(x) = \frac{2x^2 + 16 - 2x^2}{(2x^2 + 16)\sqrt{2x^2 + 16}}$

$f^{\prime \prime}(x) = \frac{16}{\sqrt{(2x^2 + 16)^3}}$

The typical ways to write this would be
$f^{\prime \prime}(x) = \frac{16}{(2x^2 + 16)^{3/2}}$
or
$f^{\prime \prime}(x) = 16(2x^2 + 16)^{-3/2}$

-Dan