# Relationship between First and Second Derivatives of a Function

## Key Questions

• Since $f ' '$ is the first derivative of $f '$, $f ' '$ tells us about the increasing/decreasing behavior of $f '$.

$2 {\sec}^{2} x \tan x$

#### Explanation:

First we find $\frac{d}{\mathrm{dx}} \tan x$.

We know that $\tan x = \sin \frac{x}{\cos} x$

So we can use the quotient rule to solve for this:

$\frac{d}{\mathrm{dx}} \left(\sin \frac{x}{\cos} x\right) = \frac{\cos x \frac{d}{\mathrm{dx}} \left(\sin x\right) - \sin x \frac{d}{\mathrm{dx}} \left(\cos x\right)}{\cos} ^ 2 x$

$\textcolor{w h i t e}{\frac{d}{\mathrm{dx}} \left(\sin \frac{x}{\cos} x\right)} = \frac{\cos x \left(\cos x\right) - \sin x \left(- \sin x\right)}{\cos} ^ 2 x$

$\textcolor{w h i t e}{\frac{d}{\mathrm{dx}} \left(\sin \frac{x}{\cos} x\right)} = \frac{{\cos}^{2} x + {\sin}^{2} x}{{\cos}^{2} x} = \frac{1}{\cos} ^ 2 x$

$\frac{d}{\mathrm{dx}} \left(\sin \frac{x}{\cos} x\right) = \frac{d}{\mathrm{dx}} \tan x = {\sec}^{2} x$

Now for ${d}^{2} / {\mathrm{dx}}^{2} \tan x$, or $\frac{d}{\mathrm{dx}} {\sec}^{2} x$

Which we can write as $\frac{d}{\mathrm{dx}} {\left(\sec x\right)}^{2}$, which gives:

$2 \sec x \left(\sec x \tan x\right)$, using the chain rule, where we compute $\frac{d}{\mathrm{du}} {u}^{2}$ and $\frac{d}{\mathrm{dx}} \sec x$.

Which gives:

$2 {\sec}^{2} x \tan x$

So:

${d}^{2} / {\mathrm{dx}}^{2} \tan x = 2 {\sec}^{2} x \tan x$

See below.

#### Explanation:

The second derivative is the derivative of the derivative of a function.

Let's take a random function, say $f \left(x\right) = {x}^{3}$. The derivative of $f \left(x\right)$, that is, $f ' \left(x\right)$, is equal to $3 {x}^{2}$.

The second derivative of ${x}^{3}$ is the derivative of $3 {x}^{2}$. That's $6 x$.

So we say that the second derivative of $f \left(x\right) = {x}^{3}$, or $f ' ' \left(x\right)$, is equal to $6 x$