## Derivative Patterns

Read this section to learn about patterns of derivatives. Work through practice problems 1-8.

The derivative of a function $\mathrm{f}$ is a new function $\mathbf{f}^{\prime}$, and we can calculate the derivative of this new function to get the derivative of the derivative of $f$, denoted by $f^{\prime \prime}$ and called the second derivative of $f$. For example, if $f(x)=x^{5}$ then $f^{\prime}(x)=5 x^{4}$ and $f^{\prime \prime}(x)=\left(f^{\prime}(x)\right)^{\prime}=\left(5 x^{4}\right)^{\prime}=20 x^{3}$.

Definitions: The first derivative of $\mathrm{f}$ is $\mathrm{f}^{\prime}(\mathrm{x})$, the rate of change of $\mathrm{f}$.

The second derivative of $f$ is $f^{\prime \prime}(x)=\left(f^{\prime}(x)\right)^{\prime}$, the rate of change of $f^{\prime}$. The third derivative of $f$ is $f^{\prime \prime \prime}(x)=\left(f^{\prime \prime}(x)\right)^{\prime}$, the rate of change of $f$ ".

For $y=f(x), f^{\prime}(x)=\frac{d y}{d x}, f^{\prime \prime}(x)=\frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d^{2} y}{d x^{2}}, f^{\prime \prime \prime}(x)=\frac{d}{d x}\left(\frac{d^{2} y}{d x^{2}}\right)=\frac{d^{3} y}{d x^{3}}$, and so on.

Practice 8: Find $\mathrm{f}^{\prime}, \mathrm{f} "$, and $\mathrm{f}^{\prime \prime \prime}$ for $\mathrm{f}(\mathrm{x})=3 \mathrm{x}^{7}, \mathrm{f}(\mathrm{x})=\sin (\mathrm{x})$, and $\mathrm{f}(\mathrm{x})=\mathrm{x} \cos (\mathrm{x})$

If $\mathrm{f}(\mathrm{x})$ represents the position of a particle at time $\mathrm{x}$, then $\mathrm{v}(\mathrm{x})=\mathrm{f}^{\prime}(\mathrm{x})$ will represent the velocity (rate of change of the position) of the particle and $\mathrm{a}(\mathrm{x})=\mathrm{v}^{\prime}(\mathrm{x})=\mathrm{f}^{\prime \prime}(\mathrm{x})$ will represent the acceleration (the rate of change of the velocity) of the particle.

Example 5: The height (feet) of a particle at time $\mathrm{t}$ seconds is $\mathrm{t}^{3}-4 \mathrm{t}^{2}+8 \mathrm{t}$. Find the height, velocity and acceleration of the particle when $\mathrm{t}=0,1$, and $2$ seconds.

Solution: $f(t)=t^{3}-4 t^{2}+8 t$ so $f(0)=0$ feet, $f(1)=5$ feet, and $f(2)=8$ feet

The velocity is $v(t)=f^{\prime}(t)=3 t^{2}-8 t+8$ so $v(0)=8 \mathrm{ft} / \mathrm{s}, v(1)=3 \mathrm{ft} / \mathrm{s}$, and $\mathrm{v}(2)=4 \mathrm{ft} / \mathrm{s}$. At each of these times the velocity is positive and the particle is moving upward, increasing in height.

The acceleration is $a(t)=6 t-8$ so $a(0)=-8 \mathrm{ft} / \mathrm{s}^{2}, a(1)=-2 \mathrm{ft} / \mathrm{s}^{2}$ and $\mathrm{a}(2)=4 \mathrm{ft} / \mathrm{s}^{2}$.

We will examine the geometric meaning of the second derivative later.