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# 6.5 Second derivative

## 6.5 Second derivative (EMCH9)

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The second derivative of a function is the derivative of the first derivative and it indicates the change in gradient of the original function. The sign of the second derivative tells us if the gradient of the original function is increasing, decreasing or remaining constant.

To determine the second derivative of the function $$f(x)$$, we differentiate $$f'(x)$$ using the rules for differentiation.

$f''(x) = \frac{d}{dx}[f'(x)]$

We also use the following notation for determining the second derivative of $$y$$:

$y'' = \frac{d}{dx}\left[\frac{dy}{dx}\right] = \frac{d^{2}y}{dx^{2}}$

## Worked example 16: Finding the second derivative

Calculate the second derivative for each of the following:

1. $$k(x) = 2x^{3} - 4x^{2} + 9$$
2. $$y = \frac{3}{x}$$
1. \begin{align*} k'(x) &= 2(3x^{2}) - 4(2x) + 0 \\ &= 6x^{2} - 8x \\ & \\ k''(x) &= 6(2x) - 8 \\ &= 12x - 8 \end{align*}
2. \begin{align*} y &= 3x^{-1} \\ & \\ \frac{dy}{dx} &= 3(-1x^{-2}) \\ &= -3x^{-2} \\ &= -\frac{3}{x^{2}} \\ & \\ \frac{d^{2}y}{dx^{2}} &= -3(-2x^{-3}) \\ &= \frac{6}{x^{3}} \end{align*}

## Second derivative

Textbook Exercise 6.6

Calculate the second derivative for each of the following:

$$g(x)=5x^{2}$$

\begin{align*} g'(x)&=10x \\ g''(x)&=10 \end{align*}

$$y=8x^{3}-7x$$

\begin{align*} \frac{dy}{dx}&=24x^{2}-7 \\ \frac{d^{2}y}{dx^{2}}&=48x \end{align*}

$$f(x)=x(x-6)+10$$

\begin{align*} f(x)&=x^{2}-6x+10 \\ f'(x)&=2x-6 \\ f''(x)&=2 \end{align*}

$$y=x^{5}-x^{3}+x-1$$

\begin{align*} \frac{dy}{dx}&=5x^{4}-3x^{2}+1 \\ \frac{d^{2}y}{dx^{2}}&=20x^{3}-6x \end{align*}

$$k(x)=(x^{2}+1)(x-1)$$

\begin{align*} k(x)&=x^{3}-x^{2}+x-1 \\ k'(x)&=3x^{2}-2x+1 \\ k''(x)&=6x-2 \end{align*}

$$p(x)=-\frac{10}{x^{2}}$$

\begin{align*} p(x)&= -10x^{-2} \\ p'(x)&=20x^{-3} = \frac{20}{x^{3}}\\ p''(x)&=-60x^{-4}=\frac{-60}{x^{4}} \end{align*}

$$q(x)=\sqrt{x}+5x^{2}$$

\begin{align*} q(x)&=x^{\frac{1}{2}}+5x^{2} \\ q'(x)&=\frac{1}{2}x^{-\frac{1}{2}}+10x \\ q''(x)&=-\frac{1}{4}x^{-\frac{3}{2}}+10 \\ &=-\frac{1}{4\sqrt{x^{3}}}+10 \end{align*}

Find the first and second derivatives of $$f(x)=5x(2x+3)$$.

\begin{align*} f(x)&=10x^{2}+15x \\ f'(x)&=20x+15\\ f''(x)&=20 \end{align*}

Find $$\dfrac{d^{2}}{dx^{2}}\left[6\sqrt[3]{x^{2}}\right]$$.

\begin{align*} y&=6\sqrt[3]{x^{2}} \\ &=6x^{\frac{2}{3}} \\ \frac{dy}{dx}&= 6\left(\frac{2}{3}\right)x^{\frac{2}{3}-1} \\ &=4x^{-\frac{1}{3}} \\ \frac{d^{2}y}{dx^{2}}&=4\left(-\frac{1}{3}\right)x^{-\frac{1}{3}-1} \\ &= -\frac{4}{3}x^{-\frac{4}{3}} \\ &=-\frac{4}{3\sqrt[3]{x^{4}}} \end{align*}

Given the function $$g: \enspace y=(1-2x)^{3}$$.

Determine $$g'$$ and $$g''$$.
\begin{align*} g(x)&=(1-2x)^{3} \\ &=(1-2x)(1-2x)^{2} \\ &=(1-2x)(1-4x+4x^{2}) \\ &=1 - 6x + 12x^{2} - 8x^{3} \\ g'(x)&=-6+24x-24x^{2} \\ g''(x)&=24-48x \end{align*}

What type of function is:

1. $$g'$$
2. $$g''$$
1. $$g'(x) =-6+24x-24x^{2} \enspace \text{ (quadratic function)} \\$$
2. $$g''(x) =24-48x \enspace \text{ (linear function)}$$
Find the value of $$g''\left(\frac{1}{2}\right)$$.
\begin{align*} g''\left(\frac{1}{2}\right)&= 24 - 48\left(\frac{1}{2}\right) \\ &= 24 - 24 \\ &=0 \end{align*}
What do you observe about the degree (highest power) of each of the derived functions?

Each derivative function is one degree lower than the previous one.