Derivatives, Properties, and Formulas

The contrapositive form of this theorem tells about some functions which do not have derivatives:

Proof of the Theorem: We assume that the hypothesis, $f$ is differentiable at the point $c$, is true so 

$\lim\limits_{h \rightarrow 0} \frac{f(c+h)-f(c)}{h}$ exists and equals $\mathrm{f}^{\prime}(\mathrm{c})$. We want to show that $\mathrm{f}$ must necessarily be continuous at $\mathrm{c}$ $: \lim\limits_{h \rightarrow 0} f(c+h)=f(c)$

Since $\mathrm{f}(\mathrm{c}+\mathrm{h})$ can be written as $f(c+h)=f(c)+\left\{\frac{f(c+h)-f(c)}{h}\right\} \cdot h$, we have

$\begin{aligned} &\lim\limits_{h \rightarrow 0} f(c+h)=\lim\limits_{h \rightarrow 0}\left(f(c)+\left\{\frac{f(c+h)-f(c)}{h}\right\} \cdot h\right) \\ &=\lim\limits_{h \rightarrow 0} f(c)+\lim\limits_{h \rightarrow 0}\left\{\frac{f(c+h)-f(c)}{h}\right\} \cdot \lim _{h \rightarrow 0}(h)=\mathrm{f}(\mathrm{c})+\mathrm{f}^{\prime}(\mathrm{c}) \cdot 0=\mathrm{f}(\mathrm{c}) \end{aligned}$

Therefore $f$ is continuous at $c$. 

It is important to clearly understand what is meant by this theorem and what is not meant: If the function is differentiable at a point, then the function is automatically continuous at that point. If the function is continuous at a point, then the function may or may not have a derivative at that point.  

If the function is not continuous at a point, then the function is not differentiable at that point. 

Example 1: Show that $\mathrm{f}(\mathrm{x})=[\mathrm{x}]=\mathrm{INT}(\mathrm{x})$ is not continuous and not differentiable at 2 (Fig. 1).

Solution: The one-sided limits, $\lim\limits_{x \rightarrow 2^{+}} \operatorname{INT}(\mathrm{x})=2$ and $\lim\limits_{x \rightarrow 2^­} \operatorname{INT}(x)=1$, have different values so $\lim\limits_{x \rightarrow 2} \operatorname{INT}(x)$ does not exist, and $\operatorname{INT}(\mathrm{x})$ is not continuous at 2. Since $\mathrm{f}(\mathrm{x})=\mathrm{INT}(\mathrm{x})$ is not continuous at 2, it is not differentiable there.

Lack of continuity is enough to imply lack of differentiability, but the next two examples show that continuity is not enough to guarantee differentiability. 

Example 2: Show that $\mathrm{f}(\mathrm{x})=|\mathrm{x}|$ is continuous but not differentiable at $\mathrm{x}=0$ (Fig. 2)

Solution: $\lim\limits_{x \rightarrow 0}|\mathrm{x}|=0=|0|$ so $\mathrm{f}$ is continuous at $0$, but we showed in Section 2.1 that the absolute value function was not differentiable at $\mathrm{x}=0$.

A function is not differentiable at a cusp or a "corner".

Example 3: Show that $\mathrm{f}(\mathrm{x})=\sqrt[3]{\mathrm{x}}=\mathrm{x}^{1 / 3}$ is continuous but not differentiable at $\mathrm{x}=0$ (Fig. 3)

Solution: $\lim\limits_{x \rightarrow 0^­} \sqrt[3]{x}=\lim\limits_{x \rightarrow 0^{+}} \sqrt[3]{x}=0$ so $\lim\limits_{x \rightarrow 0} \sqrt[3]{x}=0=\sqrt{0}$, and $\mathrm{f}$ is continuous at $0$. $\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{3} \mathrm{x}^{-(2 / 3)}=\frac{1}{3 \mathrm{x}^{2 / 3}}$ which is undefined at $\mathrm{x}=0$ so $\mathrm{f}$ is not differentiable at $0$.

A function is not differentiable where its tangent line is vertical.

Practice 1: At which integer values of $x$ is the graph of f in Fig. 4 continuous?  differentiable? 

Graphically, a function is continuous if and only if its graph is connected and does not have any holes or breaks. 

Graphically, a function is differentiable if and only if it is continuous and its graph is smooth with no  corners or vertical tangent lines. 

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 $D( 3g(x) – f(x) )$ 

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. 

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

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

You definitely need to memorize the differentiation rules, but it is vitally important that you also know how to use them. Sometimes it is clear that the function we want to differentiate is a sum or product of two obvious functions, but we commonly need to differentiate functions which involve several operations and functions. Memorizing the differentiation rules is only the first step in learning to use them. 

Example 7: Calculate $\mathbf{D}\left(\mathrm{x}^{5}+\mathrm{x} \cdot \sin (\mathrm{x})\right)$

Solution: This function is more difficult because it involves both an addition and a multiplication. Which rule(s) should we use, or, more importantly, which rule should we use first?

$\mathbf{D}\left(\mathrm{x}^{5}+\mathrm{x} \cdot \sin (\mathrm{x})\right) \quad=\mathrm{D}\left(\mathrm{x}^{5}\right)+\mathbf{D}(\mathrm{x} \cdot \sin (\mathrm{x}))$

applying the Sum Rule and trading one derivative for two easier ones

$=5 x^{4}+\{x \cdot D(\sin (x))+\sin (x) \cdot D(x)\}$ applying the product rule to $D( x.sin(x) )$ 

$=5x4 + x.cos(x) + sin(x)$ this expression has no more derivatives so we are done.

If you were evaluating the function $x^5 + x sin(x)$ for some particular value of $x$, you would (1) raise $x$ to the 5th power, (2) calculate $sin(x)$, (3) multiply $sin(x)$ by $x$, and (4) your FINAL evaluation step, SUM the values of  $x^5$ and $x sin(x)$.  

The FINAL step of your evaluation of $f$ indicates the FIRST rule to use to calculate the derivative of $f$. 

Practice 8: Which differentiation rule should you apply FIRST for each of the following: 

(a) $x \cdot \cos (x)-x^{3} \cdot \sin (x)$

(b) $(2 x-3) \cdot \cos (x)$

(c) $2 \cos (\mathrm{x})-7 \mathrm{x}^{2}$

(d) $\frac{\cos (x)+3 x}{\sqrt{x}}$

Practice 9: Calculate $D( \left.\frac{\mathrm{x}^{2}-5}{\sin (\mathrm{x})}\right)$ and $\frac{\mathbf{d}}{\mathbf{d t}}\left(\frac{\mathrm{t}^{2}-5}{\mathrm{t} \cdot \sin (\mathrm{t})}\right)$.

Example 8: A weight attached to a spring is oscillating up and down.

Over a period of time, the motion becomes "damped" because of friction and air resistance (Fig. 5), and its height at time $\mathrm{t}$ seconds is $\mathrm{h}(\mathrm{t})=5+\frac{\sin (\mathrm{t})}{1+\mathrm{t}}$ feet

What are the height and velocity of the weight after 2 seconds?

Solution: The height is

$\mathrm{h}(2)=5+\frac{\sin (2)}{1+2}=5+\frac{.909}{3}=5.303$ feet above the ground.

The velocity is $\mathrm{h}^{\prime}(2)$.

$\mathrm{h}^{\prime}(\mathrm{t})=0+\frac{(1+\mathrm{t}) \cdot \mathbf{D}(\sin (\mathrm{t}))-\sin (\mathrm{t}) \cdot \mathbf{D}(1+\mathrm{t})}{(1+\mathrm{t})^{2}}=\frac{(1+\mathrm{t}) \cdot \cos (\mathrm{t})-\sin (\mathrm{t})}{(1+\mathrm{t})^{2}}$

so $h^{\prime}(2)=\frac{3 \cos (2)-\sin (2)}{9}=\frac{-2.158}{9} \approx-0.24$ feet per second.

Practice 10: What are the height and velocity of the weight in the previous example after 5 seconds? What are the height and velocity of the weight be after a "long time"? 

Example 9: Calculate $D( x.sin(x).cos(x) )$. 

Solution: Clearly we need to use the Product Rule since the only operation in this function is multiplication, but the Product Rule deals with a product of two functions and we have the product of three; $x$ and $sin(x)$ and $cos(x)$. However, if we think of our two functions as $f(x) = x.sin(x)$ and $g(x) = cos(x)$, then we do have the product of two functions and 

$D( x.sin(x).cos(x) ) = D( f(x).g(x) ) = f(x).D( g(x) ) + g(x).D( f(x) )$

We are not done, but we have traded one hard derivative for two easier ones. We know that $D( cos(x) ) = –sin(x)$, and we can use the Product Rule (again) to calculate $D( x sin(x) )$. Then the last line of our calculation becomes 

$= –x sin2(x) + cos(x) { x cos(x) + sin(x) (1)} = –x sin2(x) + x cos2(x) + cos(x) sin(x)$. 

The derivative of a function $f$ is a new function $f '(x)$ which gives the slope of the line tangent to the graph of $f$ at each point $x$. To find the slope of the tangent line at a particular point $( c, f(c) )$ on the graph of $f$, we should first calculate the derivative $f '(x)$ and then evaluate the function $f '(x)$ at the point $x = c$ to get the number $f '(c)$. If you mistakenly evaluate $f$ first, you get a number $f(c)$, and the derivative of a constant is always equal to 0. 

Example 10: Determine the slope of the line tangent to $f(x) = 3x + sin(x)$ at $(0, f(0) )$ and $(1, f(1 ))$: 

Solution:$f '(x) = D( 3x + sin(x) ) = D(3x) + D( sin(x) ) = 3 + cos(x)$ When $x = 0$, the graph of $y = 3x + sin(x)$ goes through the point $( 0, 3(0)+sin(0) )$ with slope $f '(0) = 3 + cos(0) = 4$. When $x = 1$, the graph goes through the point $( 1, 3(1)+sin(1) ) = (1, 3.84)$ with slope $f '(1) = 3 + cos(1) ≈ 3.54$. 

Practice 11: Where do $f(x) = x^2 – 10x + 3$ and $g(x) = x^3 – 12x$ have a horizontal tangent lines ? 

Differentiability and Continuity: If a function is differentiable then it must be continuous. 

If a function is not continuous then it cannot be differentiable. 

A function may be continuous at a point and not differentiable there. 

Graphically: CONTINUOUS means connected.  

DIFFERENTIABLE means continuous, smooth and not vertical. 

Differentiation Patterns: 

$\begin{aligned}D(k f(x)) &=k \cdot D(f(x)) \\D(f+g) &=D f+D g \\D(f-g) &=\text { Df }-\mathbf{D} g \\\mathbf{D}(f \cdot g) &=f \cdot D g+g \cdot \mathbf{D} f \\D(f / g) &=\frac{g \cdot D f-f \cdot D g}{g^{2}}\end{aligned}$

The FINAL STEP used to evaluate f indicates the FIRST RULE to use to differentiate $f$.

To evaluate a derivative at a point, first differentiate and then evaluate. 

Practice 1: $\mathrm{f}$ is continuous at $\mathrm{x}=-1,0,2,4,6$, and $7 . \mathrm{f}$ is differentiable at $\mathrm{x}=-1,2,4$, and $7$

Practice 2: $\quad \mathrm{f}(\mathrm{x})=6 \mathrm{x}+8$ and $\mathrm{g}(\mathrm{x})=2$ so $\mathrm{D}(\mathrm{f}(\mathrm{x}))=6$ and $\mathrm{D}(\mathrm{g}(\mathrm{x}))=0$.

$\begin{aligned} &D(3 f(x))=3 D(f(x))=3(6)=18, \quad D(x)+g(x))=D(f(x))+D(g(x))=6+0=6 \\ &D(f(x)-g(x))=D(f(x))-D(g(x))=6-0=6 \\ &D(f(x) g(x))=f(x) D(g(x))+g(x) D(f(x))=(6 x+8)(0)+(2)(6)=12 \\ &D(f(x) / g(x))=\frac{g(x) D(f(x))-f(x) D(g(x))}{(g(x))^{2}}=\frac{(2)(6)-(6 x+8)(0)}{2^{2}}=\frac{12}{4}=3 \end{aligned}$

Practice 3: $\quad D\left(x^{3}-5 \sin (x)\right)=D\left(x^{3}\right)-5 D(\sin (x))=3 x^{2}-5 \cos (x)$

$\quad  \frac{\mathbf{d}}{\mathbf{d x}}\left(\sin (x)-4 x^{3}\right)=\frac{\mathbf{d}}{d x} \sin (x)-4 \frac{d}{d x} x^{3}=\cos (x)-12 \mathbf{x}^{2}$

Practice 4:

Practice 5: 

$\begin{align*} \begin{aligned} D(f(x)+g(x))=& \lim\limits_{h \rightarrow 0} \frac{\{f(x+h)+g(x+h)\}-\{f(x)+g(x)\}}{h} \\ &=\lim\limits_{h \rightarrow 0} \frac{f(x+h)-f(x)+g(x+h)-g(x)}{h} \\ &=\left(\lim\limits_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\right)+\left(\lim\limits_{h \rightarrow 0} \frac{g(x+h)-g(x)}{h}\right)=\mathbf{D}(\mathrm{f}(\mathrm{x}))+\mathbf{D}(\mathrm{g}(\mathrm{x})) \end{aligned} \end{align*}$

Practice 6:

$\begin{array}{l|l|l|l|c|c|c|c} x & \mathrm{f}(\mathrm{x}) & \mathrm{f}^{\prime}(\mathrm{x}) & \mathrm{g}(\mathrm{x}) & \mathrm{g}^{\prime}(\mathrm{x}) & \mathrm{D}(\mathrm{f}(\mathrm{x}) \cdot \mathrm{g}(\mathrm{x})) & \mathrm{D}(\mathrm{f}(\mathrm{x}) / \mathrm{g}(\mathrm{x})) & \mathrm{D}(\mathrm{g}(\mathrm{x}) / \mathrm{f}(\mathrm{x})) \\ \hline 0 & 3 & -2 & -4 & 3 & 3 \cdot 3+(-4)(-2)=\mathbf{1} 7 & \frac{-4(-2)-(3)(3)}{(-4)^{2}}=\frac{-\mathbf{1}}{\mathbf{1 6}} & \frac{(3)(3)-(-4)(-2)}{3^{2}}=\frac{1}{\mathbf{9}} \\ 1 & 2 & -1 & 1 & 0 & 2 \cdot 0+1(-1)=-\mathbf{1} & \frac{1(-1)-(2)(0)}{1^{2}}=-\mathbf{1} & \frac{2(0)-1(-1)}{2^{2}}=\frac{1}{4} \\ 2 & 4 & 2 & 3 & 1 & 4 \cdot 1+3 \cdot 2=\mathbf{1 0} & \frac{3(2)-(4)(1)}{3^{2}}=\frac{2}{9} & \frac{4(1)-3(2)}{4^{2}}=\frac{-\mathbf{2}}{\mathbf{1 6}} \end{array}$

Practice 7:

$\left.\mathbf{D}\left(\left(x^{2}+1\right)(7 x-3)\right)=\left(x^{2}+1\right) \mathbf{D}(7 x-3)+(7 x-3)\right)\left(x^{2}+1\right)=\left(x^{2}+1\right)(7)+(7 x-3)(2 x)=21 x^{2}-6 x+7$

or $D\left(\left(x^{2}+1\right)(7 x-3)\right)=D\left(7 x^{3}-3 x^{2}+7 x\right)=21 x^{2}-6 x+7$

$\begin{aligned}&\begin{array}{l}\frac{\text { d }}{\text { dt }}\left(\frac{3 t-2}{5 t+1}\right)=\frac{(5 t+1) D(3 t-2)-(3 t-2) D(5 t+1)}{(5 t+1)^{2}}=\frac{(5 t+1)(3)-(3 t-2)(5)}{(5 t+1)^{2}}=\frac{13}{(5 t+1)^{2}} \\ D\left(\frac{\cos (x)}{x}\right)=\frac{x D(\cos (x))-\cos (x) D(x)}{(x)^{2}}=\frac{x(-\sin (x))-\cos (x)(1)}{x^{2}}=\frac{-x \cdot \sin (x)-\cos (x)}{x^{2}} \end{array} \end{aligned}$

Practice 8: (a) difference rule (b) product rule (c) difference rule (d) quotient rule

Practice 9:

$\begin{aligned} &\mathbf{D}\left(\frac{x^{2}-5}{\sin (x)}\right)=\frac{\sin (x) D\left(x^{2}-5\right)-\left(x^{2}-5\right) D(\sin (x))}{(\sin (x))^{2}}=\frac{\sin (x)(2 x)-\left(x^{2}-5\right) \cos (x)}{\sin ^{2}(x)} \\ &\frac{\mathbf{d}}{\mathbf{d t}}\left(\frac{t^{2}-5}{t \cdot \sin (t)}\right)=\frac{t \cdot \sin (t) D\left(t^{2}-5\right)-\left(t^{2}-5\right) \mathbf{D}(t \cdot \sin (t))}{(t \cdot \sin (t))^{2}}=\frac{t \cdot \sin (t)(2 t)-\left(t^{2}-5\right)\{\mathbf{t} \cdot \cos (t)+\sin (\mathbf{t})\}}{t^{2} \cdot \sin ^{2}(t)} \end{aligned}$

Practice 10:

(a) $\quad \mathrm{h}(5)=5+\frac{\sin (5)}{1+5} \approx 4.84$ ft. $\mathrm{v}(5)=\mathrm{h}^{\prime}(5)=\frac{(1+5) \cos (5)-\sin (5)}{(1+5)^{2}} \approx 0.074 \mathrm{ft} / \mathrm{sec} .$

"long time": $\quad \lim\limits_{t \rightarrow \infty} h(t)=\lim\limits_{t \rightarrow \infty} 5+\frac{\sin (t)}{1+t}=5$ feet.

$\lim\limits_{t \rightarrow \infty} h^{\prime}(t)=\lim\limits_{t \rightarrow \infty} \frac{(1+t) \cos (t)-\sin (t)}{(1+t)^{2}}=\lim\limits_{t \rightarrow \infty}\left\{\frac{\cos (t)}{1+t}-\frac{\sin (t)}{(1+t)^{2}}\right\}=0 \mathbf{f t} / \mathbf{s e c}$

Practice 11:

$f^{\prime}(x)=2 x-10 . f^{\prime}(x)=0$ when $2 x-10=0$ so when $x=5$.

$g^{\prime}(x)=3 x^{2}-12 . g^{\prime}(x)=0$ when $3 x^{2}-12=0$ so $x^{2}=4$ and $x=-2,+2$