## Content

### The second derivative

Given a function $$f(x)$$, we can differentiate it to obtain $$f'(x)$$. It can be useful for many purposes to differentiate again and consider the second derivative of a function.

In functional notation, the second derivative is denoted by $$f''(x)$$. In Leibniz notation, letting $$y=f(x)$$, the second derivative is denoted by $$\dfrac{d^2 y}{dx^2}$$.

The placement of the $$2$$'s in the notation $$\dfrac{d^2 y}{dx^2}$$ may appear unusual. We consider that we have applied the differentiation operator $$\dfrac{d}{dx}$$ twice to $$y$$:

$\Bigl(\dfrac{d}{dx}\Bigr)^2 y = \dfrac{d^2}{dx^2}\,y = \dfrac{d^2 y}{dx^2};$

or that we have applied the differentiation operator $$\dfrac{d}{dx}$$ to $$\dfrac{dy}{dx}$$:

$\dfrac{d^2 y}{dx^2} = \dfrac{d}{dx}\Bigl(\dfrac{dy}{dx}\Bigr).$

As we will see in the module Applications of differentiation , the second derivative can be very useful in curve-sketching. The second derivative determines the convexity of the graph $$y=f(x)$$ and can, for example, be used to distinguish maxima from minima.

The second derivative can also have a physical meaning. For example, if $$x(t)$$ gives position at time $$t$$, then $$x'(t)$$ is the velocity and the second derivative $$x''(t)$$ is the acceleration at time $$t$$. This is discussed in the module Motion in a straight line .

#### Example

Find the second derivative of $$f(x) = x^2$$.

#### Solution

We have $$f'(x) = 2x$$, and so $$f''(x) = 2$$.

This example implies that, if you ride your bike and your position is $$x^2$$ metres after $$x$$ seconds, then your acceleration is a constant 2 m/s$$^2$$.

#### Example

Let $$y = x^7 + 3x^5 + x^{\dfrac{3}{2}}$$. Find $$\dfrac{d^2 y}{dx^2}$$.

#### Solution

The first derivative is

$\dfrac{dy}{dx} = 7x^6 + 15 x^4 + \dfrac{3}{2} x^{\frac{1}{2}},$

so the second derivative is

$\dfrac{d^2 y}{dx^2} = 42 x^5 + 60 x^3 + \dfrac{3}{4} x^{-\frac{1}{2}}.$
##### Exercise 18

Let $$f(x) = (x^2 + 7)^{100}$$, as in exercise 11. What is $$f''(x)$$?

##### Exercise 19

Suppose the position of an object at time $$t$$ is given by

$x(t) = 1 - 7t + (t-5)^4.$

Show that $$x''(t) \geq 0$$ for all $$t$$, so that acceleration is always non-negative.

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