Derivative Patterns

If we apply the Product Rule to the product of a function with itself, a familiar pattern emerges. 

$\begin{align*}\begin{aligned}&D\left(f^{2}\right)=D(f \cdot f)=f \cdot D(f)+f \cdot D(f)=2 f \cdot D(f) \\&D\left(f^{3}\right)=D\left(f^{2} \cdot f\right)=f^{2} \cdot \mathbf{D}(f)+f \cdot D\left(f^{2}\right)=f^{2} \cdot \mathbf{D}(f)+f\{2 f \cdot D(f)\}=f^{2} \cdot \mathbf{D}(f)+2 f^{2} \cdot \mathbf{D}(f)=3 f^{2} \cdot \mathbf{D}(f) \\&D\left(f^{4}\right)=D\left(f^{3} \cdot f\right)=f^{3} \cdot \mathbf{D}(f)+f \cdot D\left(f^{3}\right)=f^{3} \cdot \mathbf{D}(f)+f\left\{3 f^{2} \cdot \mathbf{D}(f)\right\}=f^{3} \cdot \mathbf{D}(f)+3 f^{3} \cdot \mathbf{D}(f)=4 f^{3} \cdot \mathbf{D}(\mathbf{f})\end{aligned}\end{align*}$

Practice 1: What is the pattern here? What do you think the results will be for $\mathbf{D}\left(\mathrm{f}^{5}\right)$ and $\mathrm{D}\left(\mathrm{f}^{13}\right)$?

We could keep differentiating higher and higher powers of $f(x)$ by writing them as products of lower powers of $f(x)$ and using the Product Rule, but the Power Rule For Functions guarantees that the pattern we just saw for the small integer powers also works for all constant powers of functions. 

Power Rule For Functions: If $\mathrm{n}$ is any constant,

then $\begin{align*}\quad \mathbf{D}\left(f^{n}(x)\right)=n f^{n-1}(x) \cdot \mathbf{D}(\mathbf{f}(x))\end{align*}$

The Power Rule for Functions is a special case of a more general theorem, the Chain Rule, which we will examine in Section 2.4. The Power Rule For Functions will be proved after the Chain Rule.

Example 1: Use the Power Rule for Functions to find

(a) $\mathbf{D}\left(\left(\mathrm{x}^{3}-5\right)^{2}\right)$

(b) $\frac{\mathbf{d}}{\mathrm{dx}}\left(\sqrt{2 \mathrm{x}+3 \mathrm{x}^{5}}\right)$

(c) $\mathbf{D}\left(\sin ^{2}(\mathrm{x})\right)=\mathbf{D}\left((\sin (\mathrm{x}))^{2}\right)$.

Solution: (a) To match the pattern of the Power Rule for $\mathbf{D}\left(\left(x^{3}-5\right)^{2}\right)$, let $f(x)=x^{3}-5$ and $n=2$.

Then $\mathbf{D}\left(\left(x^{3}-5\right)^{2}\right)=\mathbf{D}\left(f^{n}(x)\right)=n f^{n-1}(x) \cdot \mathbf{D}(f(x))$

$\begin{align*}=2\left(x^{3}-5\right)^{1} D\left(x^{3}-5\right)=2\left(x^{3}-5\right)\left(3 x^{2}\right)=6 x^{2}\left(x^{3}-5\right)\end{align*}$

(b) To match the pattern for $\frac{\mathbf{d}}{\mathrm{dx}}\left(\sqrt{2 \mathrm{x}+3 \mathrm{x}^{5}}\right)=\frac{\mathbf{d}}{\mathbf{d x}}\left(\left(2 \mathrm{x}+3 \mathrm{x}^{5}\right)^{1 / 2}\right)$, we can let $\mathrm{f}(\mathrm{x})=2 \mathrm{x}+3 \mathrm{x}^{5}$ and take $\mathrm{n}=1 / 2$. Then

$\begin{align*}\begin{aligned}\frac{\mathbf{d}}{\mathbf{d x}}\left(\left(2 \mathrm{x}+3 \mathrm{x}^{5}\right)^{1 / 2}\right) &=\frac{\mathbf{d}}{d \mathrm{x}}\left(\mathrm{f}^{\mathrm{n}}(\mathrm{x})\right)=\mathrm{n} \mathrm{f}^{\mathrm{n}-1}(\mathrm{x}) \cdot \frac{\mathbf{d}}{\mathrm{dx}}(\mathrm{f}(\mathrm{x}))=\frac{1}{2}\left(2 \mathrm{x}+3 \mathrm{x}^{5}\right)^{-1 / 2} \frac{\mathbf{d}}{\mathrm{dx}}\left(2 \mathrm{x}+3 \mathrm{x}^{5}\right) \\&=\frac{1}{2}\left(2 \mathrm{x}+3 \mathrm{x}^{5}\right)^{-1 / 2}\left(2+15 \mathrm{x}^{4}\right)=\frac{2+15 \mathrm{x}^{4}}{2 \sqrt{2 \mathrm{x}+3 \mathrm{x}^{5}}}\end{aligned}\end{align*}$

(c) To match the pattern for $\mathbf{D}\left(\sin ^{2}(x)\right)$, Let $f(x)=\sin (x)$ and $n=2$. Then

$\begin{align*}\mathbf{D}\left(\sin ^{2}(\mathrm{x})\right) \quad=\mathbf{D}\left(\mathrm{f}^{\mathrm{n}}(\mathrm{x})\right) \quad=\mathrm{n} \mathrm{f}^{\mathrm{n}-1}(\mathrm{x}) \cdot \mathbf{D}(\mathrm{f}(\mathrm{x}))=2 \sin ^{1}(\mathrm{x}) \mathbf{D}(\sin (\mathrm{x}))=2 \sin (\mathbf{x}) \cos (\mathbf{x})\end{align*}$

Practice 2: Use the Power Rule for Functions to find

(a) $\frac{\mathbf{d}}{\mathbf{d x}}\left(\left(2 \mathrm{x}^{5}-\pi\right)^{2}\right)$,

(b) $\mathbf{D}\left(\sqrt{x+7 x^{2}}\right)$

(c) $\mathbf{D}\left(\cos ^{4}(x)\right)=\mathbf{D}\left((\cos (x))^{4}\right)$.

Example 2: Use calculus to show that the line tangent to the circle $x^{2}+y^{2}=25$ at the point $(3,4)$ has slope $-3 / 4$.

Solution: The top half of the circle is the graph of $y=f(x)=\sqrt{25-x^{2}}$ so $f^{\prime}(x)=D\left(\left(25-x^{2}\right)^{1 / 2}\right)$

$\begin{align*}=\frac{1}{2}\left(25-\mathrm{x}^{2}\right)^{-1 / 2} \mathrm{D}\left(25-\mathrm{x}^{2}\right)=\frac{-\mathrm{x}}{\sqrt{25-\mathrm{x}^{2}}} \end{align*}$ and $\quad\mathrm{f}^{\prime}(3)=\frac{-3}{\sqrt{25-3^{2}}}=\frac{-3}{4}$

As a check, you can verify that the slope of the radial line through the center of the circle $(0,0)$ and the point $(3,4)$ has slope $4 / 3$ and is perpendicular to the tangent line which has a slope of $-3 / 4$.