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 practice 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 $f(x)=(x^2-1)^5$.
\( \begin{align} \displaystyle
f^{\prime}(x) &= \left[(x^2-1)^5\right]’ \\
&= 5(x^2-1)^{5-1} \times (x^2-1)’ \\
&= 5(x^2-1)^{4} \times 2x \\
&= 10x(x^2-1)^{4} \\
f^{\prime \prime}(x) &= \left[10x(x^2-1)^{4}\right]’ \\
&= 40x(x^2-1)^{4-1} \times (x^2-1)’ \\
&= 40x(x^2-1)^{3} \times 2x \\
&= 80x^2(x^2-1)^{3}
\end{align} \)
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 $(\cos{x})^{\prime}=-\sin{x}$.
\( \begin{align} \displaystyle
f^{\prime}(x) &= \left[\sin{(2x)}\right]’ \\
&= 2\cos{(2x)} \\
f^{\prime \prime}(x) &= \left[2\cos{(2x)})\right]’ \\
&= -4\sin{(2x)} \\
f^{(3)} &= \left[-4\sin{(2x)}\right]’ \\
&= -8\cos{(2x)}
\end{align} \)
Example 4
Find $\dfrac{d^{(3)}y}{dx^{(3)}}$ if $y=e^{2x}$, given that $\dfrac{d}{dx}e^x = e^x$.
\( \begin{align} \displaystyle
\dfrac{dy}{dx} &= e^{2x} \times \dfrac{d}{dx}2x \\
&= e^{2x} \times 2 \\
&= 2e^{2x} \\
\dfrac{d^{2}y}{dx^{2}} &= 2e^{2x} \times \dfrac{d}{dx}2x \\
&= 2e^{2x} \times 2 \\
&= 4e^{2x} \\
\dfrac{d^{3}y}{dx^{3}} &= 4e^{2x} \times \dfrac{d}{dx}2x \\
&= 4e^{2x} \times 2 \\
&= 8e^{2x}
\end{align} \)
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