## Derivatives, Properties, and Formulas

Read this section to understand the properties of derivatives. Work through practice problems 1-11.

Example 4: The derivative of $\mathrm{f}(\mathrm{x})=\mathrm{x}$ is $\mathbf{D} \mathrm{f}(\mathrm{x})=1$, and the derivative of $\mathrm{g}(\mathrm{x})=5$ is $\mathbf{D g}(\mathrm{x})=0$

What are the derivatives of their elementary combinations: $3 \mathrm{f}, \mathrm{f}+\mathrm{g}, \mathrm{f}-\mathrm{g}, \mathrm{f} \cdot \mathrm{g}$ and $\mathrm{f} / \mathrm{g}$?

Solution:

\begin{aligned} &D(3 f(x))=D(3 x)=3=3 \mathbf{D}(f(x)) \\ &D(f(x)+g(x))=D(x+5)=1=D(f(x))+D(g(x)) \\ &D(f(x)-g(x))=D(x-5)=1=D(f(x))-D(g(x)) \end{aligned}

Unfortunately, the derivatives of $\mathrm{f} \cdot \mathrm{g}$ and $\mathrm{f} / \mathrm{g}$ don't follow the same easy patterns:

$\mathbf{D}(\mathrm{f}(\mathrm{x}) \cdot \mathrm{g}(\mathrm{x}))=\mathrm{D}(5 \mathrm{x})=5$ but $\mathrm{D}(\mathrm{f}(\mathrm{x})) \cdot \mathbf{D}(\mathrm{g}(\mathrm{x}))=(1) \cdot(0)=0$, and

$D(f(x) / g(x))=D(x / 5)=1 / 5$ but $D(f(x)) / D(g(x))$ is undefined

These two very simple functions show that, in general, $\mathbf{D}(\mathrm{f} \cdot \mathrm{g}) \neq \mathbf{D}(\mathrm{f}) \cdot \mathbf{D}(\mathrm{g})$ and $\mathrm{D}(\mathrm{f} / \mathrm{g}) \neq \mathrm{D}(\mathrm{f}) / \mathrm{D}(\mathrm{g})$.

The Main Differentiation Theorem below states the correct patterns for differentiating products and quotients.

Practice 2: For $f(x) = 6x + 8$ and $g(x) = 2$, what are the derivatives of  $3f, f+g, f–g, f. g$ and $f/g$?

The following theorem says that the simple patterns in the example for constant multiples of functions and sums and differences of functions are true for all differentiable functions. It also includes the correct patterns for derivatives of products and quotients of differentiable functions.

Main Differentiation Theorem: If $f$ and $g$ are differentiable at $x$, then

(a) Constant Multiple Rule: $\quad D(k f(x))=k D(f(x))=kDf$ or $(\mathrm{kf}(\mathrm{x}))^{\prime}=\mathrm{k} \mathrm{f}^{\prime}(\mathrm{x})$

(b) Sum Rule: $\mathbf{D}(\mathrm{f}(\mathrm{x})+g(\mathrm{x}))=\mathrm{D}(\mathrm{f}(\mathrm{x}))+\mathrm{D}(\mathrm{g}(\mathrm{x}))=\mathrm{D} \mathbf{f}+\mathrm{D} \mathrm{g}$ $\quad$ or $(\mathrm{f}(\mathrm{x})+\mathrm{g}(\mathrm{x}))^{\prime}=\mathrm{f}^{\prime}(\mathrm{x})+\mathrm{g}^{\prime}(\mathrm{x})$

(c) Difference Rule: $\mathrm{D}(\mathrm{f}(\mathrm{x})-\mathrm{g}(\mathrm{x}))=\mathrm{D}(\mathrm{f}(\mathrm{x}))-\mathrm{D}(\mathrm{g}(\mathrm{x}))=\mathrm{D} \mathbf{f}-\mathrm{D} g$ $\quad$ or $(\mathrm{f}(\mathrm{x})-\mathrm{g}(\mathrm{x}))^{\prime}=\mathrm{f}^{\prime}(\mathrm{x})-\mathrm{g}^{\prime}(\mathrm{x})$

(d) Product Rule: $\quad D(f(x) \cdot g(x))=f(x) \cdot D(g(x))+g(x) \cdot D(f(x))=f D g+g Df$ or $(f(x) \cdot g(x))^{\prime}=f(x) \cdot g^{\prime}(x)+g(x) \cdot f^{\prime}(x)$

(e) Quotient Rule: $D\left(\frac{f(x)}{g(x)}\right)=\frac{g(x) \cdot D(f(x))-f(x) \cdot D(g(x))}{(g(x))^{2}}$

=$\frac{\mathrm{g} \mathbf{D f}-\mathrm{f} \mathbf{D} \mathbf{g}}{\mathrm{g}^{2}}$ or $\frac{g(x) \cdot f^{\prime}(x)-f(x) \cdot g^{\prime}(x)}{g^{2}(x)}$

(provided $g(x) ≠ 0$)

The proofs of parts (a), (b), and (c) of this theorem are straightforward, but parts (d) and (e) require some clever algebraic manipulations. Lets look at an example first.

Example 5: Recall that $\mathbf{D}\left(\mathrm{x}^{2}\right)=2 \mathrm{x}$ and $\mathbf{D}(\sin (\mathrm{x}))=\cos (\mathrm{x})$. Find $\mathbf{D}(3 \sin (\mathrm{x}))$ and $\frac{\mathbf{d}}{\mathbf{d x}}\left(5 \mathrm{x}^{2}-7 \sin (\mathrm{x})\right)$.

Solution: $\mathbf{D}(3 \sin (\mathrm{x}))$ is an application of part (a) of the theorem with $\mathrm{k}=3$ and $\mathrm{f}(\mathrm{x})=\sin (\mathrm{x})$ so $\mathbf{D}(3 \sin (x))=3 \mathbf{D}(\sin (x))=3 \cos (x)$

$\frac{d}{d x}\left(5 x^{2}-7 \sin (x)\right)$ uses part (c) of the theorem with $f(x)=5 x^{2}$ and $g(x)=7 \sin (x)$ so

\begin{aligned} \frac{\mathbf{d}}{\mathbf{d x}}\left(5 \mathrm{x}^{2}-7 \sin (\mathrm{x})\right)=\frac{\mathbf{d}}{\mathbf{d x}}\left(5 \mathrm{x}^{2}\right)-\frac{\mathbf{d}}{\mathbf{d x}}(7 \sin (\mathrm{x})) &=5 \frac{\mathbf{d}}{\mathbf{d x}}\left(\mathrm{x}^{2}\right)-7 \frac{\mathbf{d}}{\mathbf{d x}}(\sin (\mathrm{x})) \\ &=5(\mathbf{2} \mathbf{x})-7(\cos (\mathbf{x}))=10 \mathrm{x}-7 \cos (\mathrm{x}) \end{aligned}

Practice 3: Find $\mathbf{D}\left(\mathrm{x}^{3}-5 \sin (\mathrm{x})\right)$ and $\frac{\mathbf{d}}{\mathbf{d x}}\left(\sin (\mathrm{x})-4 \mathrm{x}^{3}\right)$.

Practice 4: Fill in the values in the table for $D( 3f(x) ), D( 2f(x)+g(x) )$, and $D( 3g(x) – f(x) )$

$x$

$f(x)$

$f '(x)$

$g(x)$

$g '(x)$

$D( 3f(x) )$

$D( 2f(x)+g(x) )$

$D( 3g(x)-f(x) )$

0

3

–2

–4

3

1

2

–1

1

0

2

4

2

3

1

Proof of the Main Derivative Theorem (a) and (c): The only general fact we have about derivatives is the definition as a limit, so our proofs here will have to recast derivatives as limits and then use some results about limits. The proofs  are applications of the definition of the derivative and results about limits.

(a) $\mathrm{D}(\mathrm{kf}(\mathrm{x})) \equiv \lim\limits_{h \rightarrow 0} \frac{k \cdot f(x+h)-k \cdot f(x)}{h}=\lim\limits_{h \rightarrow 0} k \cdot \frac{f(x+h)-f(x)}{h}=k \cdot \lim\limits_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=\mathrm{k} \cdot \mathbf{D}(\mathrm{f}(\mathrm{x}))$.

(c) $\mathbf{D}(\mathrm{f}(\mathrm{x})-\mathrm{g}(\mathrm{x}))=\lim\limits_{h \rightarrow 0} \frac{\{f(x+h)-g(x+h)\}-\{f(x)-g(x)\}}{h}$

$\lim\limits_{h \rightarrow 0} \frac{\{f(x+h)-f(x)\}-\{g(x+h)-g(x)\}}{h}=\lim\limits_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=\lim\limits_{h \rightarrow 0} \frac{g(x+h)-g(x)}{h}$

$=\mathbf{D}(\mathrm{f}(\mathrm{x}))-\mathbf{D}(\mathrm{g}(\mathrm{x}))$

The proof of part (b) is very similar to these two proofs, and is left for you as the next Practice Problem.

The proof for the Product Rule and Quotient Rules will be given later.

Practice 5: Prove part (b) of the theorem, the Sum Rule: $D( f(x) + g(x) ) = D( f(x) ) + D( g(x) )$

Practice 6: Use the Main Differentiation Theorem and the values in the table to fill in the rest of the table.

$x$

$f(x)$

$f '(x)$

$g(x)$

$g '(x)$

$D( f(x).g(x) )$

$D( f(x)/g(x) )$

$D( g(x)/f(x)$

0

3

–2

–4

3

1

2

–1

1

0

2

4

2

3

1

Example 6: Determine $\mathbf{D}\left(\mathrm{x}^{2} \cdot \sin (\mathrm{x})\right)$ and $\frac{\mathbf{d}}{\mathbf{d x}}\left(\frac{\mathrm{x}^{3}}{\sin (\mathrm{x})}\right)$.

Solution: (a) We can use the Product Rule with $\mathrm{f}(\mathrm{x})=\mathrm{x}^{2}$ and $\mathrm{g}(\mathrm{x})=\sin (\mathrm{x})$:

\begin{aligned} \mathbf{D}\left(x^{2} \cdot \sin (x)\right)=D(f(x) \cdot g(x)) &=f(x) \cdot D(g(x))+g(x) \cdot D(f(x)) \\ &=\left(x^{2}\right) \cdot \mathbf{D}(\sin (x))+\sin (x) \cdot \mathbf{D}\left(x^{2}\right) \\ &=\left(x^{2}\right) \cdot(\cos (x))+\sin (x) \cdot(2 x)=x^{2} \cdot \cos (x)+2 x \cdot \sin (x) \end{aligned}

(b) We can use the Quotient Rule with $\mathrm{f}(\mathrm{x})=\mathrm{x}^{3}$ and $\mathrm{g}(\mathrm{x})=\sin (\mathrm{x})$:

\begin{aligned} \frac{\mathbf{d}}{\mathbf{d x}}\left(\frac{x^{3}}{\sin (x)}\right) &=\frac{g(x) \cdot D(f(x))-f(x) \cdot D(g(x))}{g^{2}(x)}=\frac{\sin (x) \cdot D\left(x^{3}\right)-x^{3} \cdot \mathbf{D}(\sin (x))}{(\sin (x))^{2}} \\ &=\frac{\sin (x) \cdot\left(3 x^{2}\right)-x^{3} \cdot \cos (x)}{\sin ^{2}(x)}=\frac{3 x^{2} \sin (x)-x^{3} \cos (x)}{\sin ^{2}(x)} \end{aligned}

Practice 7: Determine $\mathbf{D}\left(\left(x^{2}+1\right)(7 x-3)\right), \frac{\mathbf{d}}{\text { dt }}\left(\frac{3 t-2}{5 t+1}\right)$ and $D( \left.\frac{\cos (x)}{x}\right)$.

Proof of the Product Rule: The proofs of parts (d) and (e) of the theorem are complicated but only  involve elementary techniques, used in just the right way. Sometimes we will omit such computational proofs, but the Product and Quotient Rules are fundamental techniques you will need hundreds of times.

By the hypothesis, $f$ and $g$ are differentiable so $\lim\limits_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=f^{\prime}(x)$ and $\lim\limits_{h \rightarrow 0} \frac{g(x+h)-g(x)}{h}=\mathrm{g}^{\prime}(x)$.

Also, both $f$ and $g$ are continuous (why?) so $\lim\limits_{h \rightarrow 0} \mathrm{f}(\mathrm{x}+\mathrm{h})=\mathrm{f}(\mathrm{x})$ and $\lim\limits_{h \rightarrow 0} \mathrm{~g}(\mathrm{x}+\mathrm{h})=\mathrm{g}(\mathrm{x})$.

(d) Product Rule: Let $\mathrm{P}(\mathrm{x})=\mathrm{f}(\mathrm{x}) \cdot \mathrm{g}(\mathrm{x})$. Then $\mathrm{P}(\mathrm{x}+\mathrm{h})=\mathrm{f}(\mathrm{x}+\mathrm{h}) \cdot \mathrm{g}(\mathrm{x}+\mathrm{h})$.

$\mathbf{D}(\mathrm{f}(\mathrm{x}) \cdot \mathrm{g}(\mathrm{x}))=\mathrm{D}(\mathrm{P}(\mathrm{x}))=\lim\limits_{h \rightarrow 0} \frac{P(x+h)-P(x)}{h}=\lim\limits_{h \rightarrow 0} \frac{f(x+h) g(x+h)-f(x) g(x)}{h}$

$=\lim\limits_{h \rightarrow 0} \frac{f(x+h) g(x+h)+\{-f(x) g(x+h)+f(x) g(x+h)\}-f(x) g(x)}{h}$ adding and subtracting $f(x)g(x+h)$

$=\lim\limits_{h \rightarrow 0} \frac{f(x+h) g(x+h)-f(x) g(x+h)}{h}+\lim\limits_{h \rightarrow 0} \frac{f(x) g(x+h)-f(x) g(x)}{h}$ regrouping the terms

 $=\lim\limits_{h \rightarrow 0}(g(x+h)) \cdot\left(\frac{\mathrm{f}(\mathrm{x}+\mathrm{h})-\mathrm{f}(\mathrm{x})}{\mathrm{h}}\right)+(\mathrm{f}(\mathrm{x})) \cdot\left(\frac{\mathrm{g}(\mathrm{x}+\mathrm{h})-\mathrm{g}(\mathrm{x})}{\mathrm{h}}\right)$ finding common factors $\qquad \qquad \quad \downarrow \qquad \qquad \quad \downarrow \qquad \qquad \quad \downarrow \qquad \qquad \downarrow \qquad$ taking limit as $\mathrm{h} \rightarrow 0$ $\qquad \qquad (g(x)) \quad \cdot \quad \left(f^{\prime}(x)\right) \, \, \, + \, \, (f(x)) \, \, \cdot \, \, \left(g^{\prime}(x)\right)=g D f+f D g$

(e) The steps for a proof of the Quotient Rule are shown in Problem 55.