# Higher Derivatives

Given a function $f(x)$, the derivative $f^{\prime}(x)$ is known as the first derivative.
The second derivative of $f(x)$ is the derivative of $f^{\prime}(x)$, which is $f^{\prime \prime}(x)$ or the derivative of the first derivative.
\displaystyle \begin{align} f^{\prime}(x) &= \dfrac{d}{dx}f(x) \\ f^{\prime \prime}(x) &= \dfrac{d}{dx}f'(x) \\ f^{(3)}(x) &= \dfrac{d}{dx}f^{\prime \prime}(x) \\ f^{(4)}(x) &= \dfrac{d}{dx}f^{(3)}(x) \\ &\cdots \\ f^{(n)}(x) &= \dfrac{d}{dx}f^{(n-1)}(x) \\ \end{align}
We can continue to differentiate to obtain higher derivatives.
The $n$th derivative if $y$ with respect to $x$ is obtained by differentiating $y=f(x)$ $n$ times. We use the notation $f^{(n)}(x)$ or $\dfrac{d^ny}{dx^n}$ for the $n$th derivative.

Let’s do some practices for this now!

### Example 1

Find $f^{\prime \prime}(x)$ given that $f(x)=x^4-3x^2-4x+5$.

\begin{align} \displaystyle f^{\prime}(x) &= (x^4-3x^2-4x+5)’ \\ &= 4x^3 – 6x – 4 \\ f^{\prime \prime}(x) &= (4x^3 – 6x – 4)’ \\ &= 12x^2 – 6 \end{align}

### Example 2

Find $f^{\prime \prime}(x)$ given that $f(x)=(x^2-1)^5$.

\begin{align} \displaystyle f^{\prime}(x) &= ((x^2-1)^5)’ \\ &= 5(x^2-1)^{5-1} \times (x^2-1)’ \\ &= 5(x^2-1)^{4} \times 2x \\ &= 10x(x^2-1)^{4} \\ f^{\prime \prime}(x) &= (10x(x^2-1)^{4})’ \\ &= 40x(x^2-1)^{4-1} \times (x^2-1)’ \\ &= 40x(x^2-1)^{3} \times 2x \\ &= 80x^2(x^2-1)^{3} \\ \end{align}

## Extension Examples

These Extension Examples require to have some prerequisite skills including;
\begin{align} \displaystyle \dfrac{d}{dx}\sin{x} &= \cos{x} \\ \dfrac{d}{dx}\cos{x} &= -\sin{x} \\ \dfrac{d}{dx}e^x &= e^x \\ \end{align}

### Example 3

Find $f^{(3)}(x)$ if $f(x)=\sin{(2x)}$, given that $(\sin{x})^{\prime}=\cos{x}$ and $(\cos{x})^{\prime}=-\sin{x}$.

\begin{align} \displaystyle f^{\prime}(x) &= (\sin{(2x)})’ \\ &= 2\cos{(2x)} \\ f^{\prime \prime}(x) &= (2\cos{(2x)})’ \\ &= -4\sin{(2x)} \\ f^{(3)} &= (-4\sin{(2x)})’ \\ &= -8\cos{(2x)} \\ \end{align}

### Example 4

Find $\dfrac{d^{(3)}y}{dx^{(3)}}$ if $y=e^{2x}$, given that $\dfrac{d}{dx}e^x = e^x$.

\begin{align} \displaystyle \dfrac{dy}{dx} &= e^{2x} \times \dfrac{d}{dx}2x \\ &= e^{2x} \times 2 \\ &= 2e^{2x} \\ \dfrac{d^{2}y}{dx^{2}} &= 2e^{2x} \times \dfrac{d}{dx}2x \\ &= 2e^{2x} \times 2 \\ &= 4e^{2x} \\ \dfrac{d^{3}y}{dx^{3}} &= 4e^{2x} \times \dfrac{d}{dx}2x \\ &= 4e^{2x} \times 2 \\ &= 8e^{2x} \end{align}