The Definition of the Derivative

In the first section of the Limits chapter, we saw that the computation of the slope of a tangent line, the instantaneous rate of change of a function, and the instantaneous velocity of an object at $x=a$ all require us to compute the following limit: $$\lim_{x\to a} \frac{f(x)-f(a)}{x-a}$$

Note that this is somtimes given the notation $$\lim_{\Delta x\to 0}\frac{\Delta y}{\Delta x}$$ where $\Delta y$ and $\Delta x$ represent the change or difference in $y$ and $x$, respectively. So $\Delta y=f(x)-f(a)$ (the difference of $y$-values or function values) and $\Delta x=x-a$ (the difference of $x$-values).

We also saw that with a small change of notation, this limit could be written as \begin{equation}\lim_{h\to 0}\frac{f(a+h)-f(a)}{h} \end{equation}

Recall that the expressions in these limits are called difference quotients. This limit is so important, and it arises in so many places, that we give it a name. It is called the derivative of the function $f.$ Here is the formal definition of the derivative.

In this definition, we read $f'(x)$ as "f prime of x." Note that we replaced all the a's in the expression with x's to acknowledge the fact that the derivative is itself a function.

Alternate Notation

Given a function $y=f(x),$ all of the following are equivalent and represent the derivative of $f(x)$ with respect to $x$:

The "fractional" notation for the derivative is called the Leibniz notation, after Gottfried Wilhelm Leibniz, one of the acknowledged inventors of calculus. This notation is convenient because it gives us a way of annotating the name of the function (the variable in the numerator) as well as the independent variable (the variable in the denominator) all in one place.

Note as well that on occasion we will drop the $(x)$ part on the function to simplify the notation somewhat. In these instances, the following are equivalent: $$f'(x) = f'.$$

Examples

Now let's use the definition to compute some derivatives.

Higher-Order Derivatives

Consider the function $f(x) = x^3 +2x -3$ from Example 2. There, we found that its derivative is $$f'(x) = 3x^2 + 2$$ Now, this is also a function, so we should be able to find its derivative as well.

The derivative we found in the last example is called the second derivative of the function $f$ and $f'(x)$ is now called the first derivative. You have probably noticed that the second derivative is again a function, and we should be able to differentiate it. This will be called the third derivative and is denoted $f'''(x) = (f''(x))'.$ You can verify, using the definition of the derivative, that $f'''(x) = 6$ in this example.

As long as the result of differentiation is again a function, we can continue this process. The next derivative is called the fourth derivative and so on. At this point, we are going to change the notation, though. The fourth derivative is denoted $$f^{(4)}(x) = (f'''(x))'$$

In general, derivatives above the third derivative will be denoted this way. We will need to be careful with this numbered notation for derivatives, though. Consider each of the following:

The presence of parentheses in the exponent denotes differentiation while the absence of parentheses denotes exponentiation.

We can also use Leibniz notation for higher-order derivatives. Given $y = f(x),$ all of the following are equivalent and represent the second derivative of $f(x)$ with respect to $x$:

The exponents in this notation indicate that we are taking the second derivative of $y$ with respect to $x$ twice. If you go on to work with functions of more than one variable, this will be a very useful way to keep track of the variables in derivatives.

Similarly we can write a general derivative as $$f^{(n)}(x) = \frac{d^n y}{dx^n}$$

As a final note, if all we are working with is the first derivative, we will often still just call it "the derivative."

Differentiability of Functions

Let's start this section with one more example. This one is a little different, but it makes a point that needs to be made.

In this last example, we saw a function for which the derivative does not exist at a point. This is a fact of life that we have to be aware of: derivatives do not always exist. Note as well that this doesn't say anything about whether or not the derivative exists anywhere else. In fact, the derivative of the absolute value function exists at every point except the one we just looked at, $x=0.$

The preceding discussion leads to the following definition:

The next theorem shows us a very nice relationship between functions that are continuous and those that are differentiable.

Another way to read the theorem above (called its contrapositive) is that if a function is not continuous, then it is not differentiable. In other words, continuity is a requirement for the derivative to exist.

Note also that the converse of this theorem is not true (that is, the theorem does not work in reverse). There are functions that are continuous at $x=a$ but not differentiable at $x=a.$

Consider $f(x) = |x|.$ We can easily see that $$\lim_{x\to 0} f(x) = \lim_{x\to 0} |x| = 0 = f(0).$$ So $f(x) = |x|$ is continuous at $x=0.$ But we just showed above in Example 4 that $f(x) = |x|$ is not differentiable at $x=0.$

The tangent line to a straight line is just the line itself. So in this graph, we can clearly see that the tangent line to any point with $x\lt 0$ has slope equal to $-1,$ while the tangent line to any point with $x \gt 0$ has slope equal to $1.$ At $(0,0),$ there is no clearly defined tangent line. The sharp corner or cusp we see on this graph is a signal that the derivative fails to exist at that point.

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Let's consider some additional situations in which a continuous function fails to be differentiable. Consider the function $f(x) = \sqrt[3]{x}.$ Its derivative at $x=0$ is

The function $f(x) = \left\{\begin{array}{ll} x\sin\left(\frac{1}{x}\right) & \text{if } x\neq 0\\ 0 & \text{if } x=0\\ \end{array}\right.$ also has a derivative that exhibits interesting behavior at $x=0.$ We see that $$f'(0) = \lim_{x\to 0} \frac{x\sin(1/x) - 0}{x-0} = \lim_{x\to 0} \sin\left(\frac{1}{x}\right).$$ This limit does not exist, essentially because the slopes of the secant lines continuously change direction as $x$ approaches zero.

Note: The following content is adapted from Paul's Online Math Notes, with permission, and is not shared under the Creative Commons license.

Summary

• We observed that if a function is not continuous, it cannot be differentiable, since every differentiable function must be continuous. However, if a function is continuous, it can still fail to be differentiable.
• We saw that $f(x)=|x|$ failed to be differentiable at $x=0$ because the limit of the slopes of the tangent lines on the left and the right were not the same. Visually, this resulted in a sharp corner on the graph of the function at $(0,0).$ From this we conclude that in order to be differentiable at a point, a function must be "smooth" at that point.
• As we saw in the example of $f(x)=\sqrt[3]{x},$ a function fails to be differentiable at a point where there is a vertical tangent line.
• As we saw with $f(x) = \left\{\begin{array}{ll} x\sin\left(\frac{1}{x}\right) & \text{if } x\neq 0\\ 0 & \text{if } x=0\\ \end{array}\right.$ a function may fail to be differentiable at a point in more complicated ways as well.

Conclusion

As a final note in this section, we will acknowledge that computing most derivatives directly from the definition is a fairly complex process filled with opportunities to make mistakes. In a couple of sections we'll start developing formulas and properties that will help us to take the derivative of many of the common functions, so that we won't need to resort to the definition of the derivative too often.

This does not mean however that it isn't important to know the definition of the derivative! There are some functions whose derivative we cannot find any other way, so it is an important definition that we should always know and keep in the back of our minds.

Practice Problems

In problems 1 - 11, use the definition of the derivative to find the first derivative of the function.

1. $f(x) = 6$

   Click for solution

   When we substitute the function into the definition of the derivative, we see that for this problem, the difference quotient simplifies really nicely:

   $$f'(x) = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h} = \lim_{h\to 0} \frac{6 - 6}{h} = \lim_{h\to 0} \frac{0}{h} = \lim_{h\to 0} (0) = 0$$

   The derivative is then $$f'(x) = 0$$

2. $V(t) = 3 - 14t$

   Click for solution

   First we put the function into the definition of the derivative.

   $$V'(t) = \lim_{h\to 0} \frac{V(t+h)-V(t)}{h} = \lim_{h\to 0} \frac{3-14(t+h)-(3-14t)}{h}$$

   Make sure that you evaluate $V(t+h)$ correctly. Not putting the "$t+h$" in the right places is one of the more common errors that people make with these problems.

   Also watch for the parentheses around the function after the subtraction. Since you are subtracting the whole function, you need to make sure that you deal with the minus sign properly, and distribute it through. This is another very common error and is one that will often make the problem impossible to complete.

   Now all we need to do is some simplification and we'll be done.

   $$V'(t) = \lim_{h\to 0} \frac{3-14t-14h-3+14t}{h} = \lim_{h\to 0}\frac{-14h}{h} = \lim_{h\to 0} (-14) = -14$$

   The derivative is then $$V'(t) = -14$$

3. $g(x) = x^2$

   Click for solution

   We first put the function into the definition of the derivative. Make sure that you properly evaluate $g(x+h).$ This is a place where people commonly make mistakes with these problems.

   All we need to do then is some quick algebra before taking the limit.

   \begin{align*} g'(x) = \lim_{h\to 0} \frac{g(x+h) - g(x)}{h} &= \lim_{h\to 0}\frac{(x+h)^2 - x^2}{h}\\ = \lim_{h\to 0} \frac{x^2+2xh+h^2-x^2}{h} &= \lim_{h\to 0} \frac{h(2x+h)}{h}\\ &= \lim_{h\to 0} (2x+h)\\ &= 2x\\ \end{align*}

   The derivative is then $$g'(x) = 2x$$

4. $Q(t) = 10 + 5t -t^2$

   Click for solution

   First we put the function into the definition of the derivative.

   $$Q'(t) = \lim_{h\to 0} \frac{Q(t+h)-Q(t)}{h} = \lim_{h\to 0} \frac{10+5(t+h)-(t+h)^2 - (10+5t-t^2)}{h}$$

   Make sure that you evaluate $Q(t+h)$ correctly. Also watch for the parentheses around the function after the subtraction. Since you are subtracting the whole function, you need to make sure that you deal with the minus sign properly, and distribute it through. This is another very common error and is one that will often make the problem impossible to complete.

   Now all we need to do is some algebra (it might be a little messy here, but that is somewhat common with these types of problems) and we'll be done.

   \begin{align*} Q'(t) &= \lim_{h\to 0} \frac{10 + 5t + 5h - t^2 - 2th - h^2 -10 -5t +t^2}{h}\\ &= \lim_{h\to 0} \frac{h(5-2t-h)}{h} = \lim_{h\to 0} (5-2t-h) = 5-2t\\ \end{align*}

   The derivative is then $$Q'(t) = 5-2t$$

5. $W(z) = 4z^2 - 9z$

   Click for solution

   First we put the function into the definition of the derivative.

   $$W'(z) = \lim_{h\to 0} \frac{W(z+h)-W(z)}{h} = \lim_{h\to 0} \frac{4(z+h)^2 - 9(z+h) - (4z^2 - 9z)}{h}$$

   Make sure that you evaluate $W(z+h)$ correctly. Also watch for the parentheses around the function after the subtraction. Since you are subtracting the whole function, you need to make sure that you deal with the minus sign properly, and distribute it through. This is another very common error and is one that will often make the problem impossible to complete.

   Now all we need to do is some algebra (it might be a little messy here, but that is somewhat common with these types of problems) and we'll be done.

   \begin{align*} W'(z) &= \lim_{h\to 0} \frac{4(z^2 + 2zh + h^2) - 9z-9h - 4z^2 +9z}{h}\\ &= \lim_{h\to 0} \frac{4z^2 + 8zh + 4h^2 - 9z - 9h - 4z^2 +9z}{h}\\ &= \lim_{h\to 0} \frac{h(8z+4h-9)}{h} = \lim_{h\to 0} (8z+4h-9) = 8z-9\\ \end{align*}

   The derivative is then $$W'(z) = 8z-9$$

6. $f(x) = 2x^3 - 1$

   Click for solution

   First we put the function into the definition of the derivative.

   $$f'(x) = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h} = \lim_{h\to 0} \frac{2(x+h)^3-1-(2x^3 - 1)}{h}$$

   Make sure that you evaluate $f(x+h)$ correctly. Also watch for the parentheses around the function after the subtraction. Since you are subtracting the whole function, you need to make sure that you deal with the minus sign properly, and distribute it through. This is another very common error and is one that will often make the problem impossible to complete.

   Now all we need to do is some algebra (it might be a little messy here, but that is somewhat common with these types of problems) and we'll be done.

   \begin{align*} f'(x) &= \lim_{h\to 0} \frac{2(x^3 + 3x^2h + 3xh^2 + h^3) -1 -2x^3 +1}{h}\\ &= \lim_{h\to 0} \frac{2x^3 + 6x^2h +6xh^2 + 2h^3 -1 -2x^3 +1}{h}\\ &= \lim_{h\to 0} \frac{h(6x^2 + 6xh +2h^2)}{h} = \lim_{h\to 0} (6x^2 + 6xh + 2h^2) = 6x^2\\ \end{align*}

   The derivative is then $$f'(x) = 6x^2$$

7. $g(x) = x^3 -2x^2 +x -1$

   Click for solution

   First we put the function into the definition of the derivative.

   $$g'(x) = \lim_{h\to 0} \frac{g(x+h)-g(x)}{h} = \lim_{h\to 0} \frac{(x+h)^3 - 2(x+h)^2 + x+h - 1 - (x^3 -2x^2 +x-1)}{h}$$

   Make sure that you evaluate $g(x+h)$ correctly. Also watch for the parentheses around the function after the subtraction. Since you are subtracting the whole function, you need to make sure that you deal with the minus sign properly, and distribute it through. This is another very common error and is one that will often make the problem impossible to complete.

   Now all we need to do is some algebra (it might be a little messy here, but that is somewhat common with these types of problems) and we'll be done.

   \begin{align*} g'(x) &= \lim_{h\to 0} \frac{x^3 +3x^2h+3xh^2 +h^3 -2(x^2+2xh+h^2)+x+h-1-(x^3-2x^2+x-1)}{h}\\ &= \lim_{h\to 0} \frac{x^3 + 3x^2h +3xh^2 + h^3 -2x^2 -4xh -2h^2 +x +h-1-x^3 +2x^2 -x+1}{h}\\ &= \lim_{h\to 0} \frac{h(3x^2+3xh+h^2-4x-2h+1)}{h}\\ &= \lim_{h\to 0} (3x^2+3xh+h^2-4x-2h+1) = 3x^2-4x+1\\ \end{align*}

   The derivative is then $$g'(x) = 3x^2-4x+1$$

8. $R(z) = \displaystyle\frac5z$

   Click for solution

   First we put the function into the definition of the derivative.

   $$R'(z) = \lim_{h\to 0} \frac{R(z+h)-R(z)}{h} = \lim_{h\to 0} \frac1h\left (\frac{5}{z+h}-\frac5z\right)$$

   Make sure that you evaluate $R(z+h)$ correctly. Also note that we wrote the overall fraction in a much more compact way. This will make it easier to see what's going on with the algebra.

   Next we need to combine the two fractions in the parentheses into a single rational expression.

   $$R'(z) = \lim_{h\to 0} \frac1h\left (\frac{5z-5(z+h)}{z(z+h)}\right)$$

   Now all we need to do is some algebra.

   $$R'(z) = \lim_{h\to 0} \frac1h\left(\frac{5z-5z-5h}{z(z+h)}\right) = \lim_{h\to 0}\frac1h\left(\frac{-5h}{z(z+h)}\right) = \lim_{h\to 0} \frac{-5}{z(z+h)} = -\frac{5}{z^2}$$

   The derivative is then $$R'(z) = -\frac{5}{z^2}$$

9. $V(t) = \displaystyle\frac{t+1}{t+4}$

   Click for solution

   First we put the function into the definition of the derivative.

   $$V'(t) = \lim_{h\to 0} \frac{V(t+h)-V(t)}{h} = \lim_{h\to 0} \frac1h\left (\frac{t+h+1}{t+h+4}-\frac{t+1}{t+4}\right)$$

   Make sure that you evaluate $V(t+h)$ correctly. Also note that we wrote the overall fraction in a much more compact way. This will make it easier to see what's going on with the algebra.

   Next we need to combine the two fractions in the parentheses into a single rational expression.

   $$V'(t) = \lim_{h\to 0} \frac1h\left (\frac{(t+h+1)(t+4)-(t+1)(t+h+4)}{(t+h+4)(t+4)}\right)$$

   Now all we need to do is some algebra (and it will get a little messy here, but that is somewhat common with these types of problems) and we'll be done.

   \begin{align*} V'(t) &=\lim_{h\to 0} \frac1h\left(\frac{t^2+th+5t+4h+4-(t^2+th+5t+h+4)}{(t+h+4)(t+4)}\right)\\ &=\lim_{h\to 0} \frac1h\left(\frac{t^2+th+5t+4h+4-t^2-th-5t-h-4}{(t+h+4)(t+4)}\right)\\ &=\lim_{h\to 0} \frac1h\left(\frac{3h}{(t+h+4)(t+4)}\right) = \lim_{h\to 0}\frac{3}{(t+h+4)(t+4)} = \frac{3}{(t+4)^2}\\ \end{align*}

   The derivative is then $$V'(t) = \frac{3}{(t+4)^2}$$

10. $Z(t) = \sqrt{3t-4}$

    Click for solution

    First we need to put the function into the definition of the derivative.

    $$Z'(t) = \lim_{h\to 0} \frac{Z(t+h)-Z(t)}{h} = \lim_{h\to 0}\frac{\sqrt{3(t+h)-4}-\sqrt{3t-4}}{h}$$

    Make sure that you evaluate $Z(t+h)$ correctly. This is a place where people commonly make mistakes with these problems.

    Next we need to rationalize the numerator.

    $$Z'(t) = \lim_{h\to 0} \frac{\left(\sqrt{3(t+h)-4}-\sqrt{3t-4}\right)}{h} \frac{\left(\sqrt{3(t+h)-4}+\sqrt{3t-4}\right)}{\left(\sqrt{3(t+h)-4}+\sqrt{3t-4}\right)}$$

    Now all that we need to do is some algebra (and it will get a little messy here, but that is pretty common with these types of problems) and we'll be done.

    \begin{align*} Z'(t) &= \lim_{h\to 0}\frac{3(t+h)-4-(3t-4)}{h\left(\sqrt{3(t+h)-4}+\sqrt{3t-4} \right)} = \lim_{h\to 0}\frac{3t+3h-4-3t+4}{h\left(\sqrt{3(t+h)-4}+\sqrt{3t-4} \right)}\\ &=\lim_{h\to 0}\frac{3h}{h\left(\sqrt{3(t+h)-4}+\sqrt{3t-4} \right)}=\lim_{h\to 0}\frac{3}{\sqrt{3(t+h)-4}+\sqrt{3t-4}}\\ &=\frac{3}{2\sqrt{3t-4}}\\ \end{align*}

    Be careful when multiplying out the numerator in the first step here. It is easy to lose track of the minus sign (or parentheses for that matter) on the second term.

    The derivative is then $$Z'(t) = \frac{3}{2\sqrt{3t-4}}$$

11. $f(x) = \sqrt{1 - 9x}$

    Click for solution

    First we put the function into the definition of the derivative: $$f'(x) = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h} = \lim_{h\to 0}\frac{\sqrt{1-9(x+h)}-\sqrt{1-9x}}{h}$$ Make sure that you evaluate $f(x+h)$ correctly. This is a place where people commonly make mistakes with these problems.

    Next we rationalize the numerator: $$f'(x) = \lim_{h\to 0} \frac{\left(\sqrt{1-9(x+h)}-\sqrt{1-9x}\right)}{h} \frac{\left(\sqrt{1-9(x+h)} +\sqrt{1-9x}\right)}{\left(\sqrt{1-9(x+h)} +\sqrt{1-9x}\right)}$$

    Now all we need to do is some algebra (which can be a little messy, but that is common with these problems).

    \begin{align*} f'(x) &= \lim_{h\to 0}\frac{1-9(x+h)-(1-9x)}{h\left(\sqrt{1-9(x+h)}+\sqrt{1-9x}\right)} = \lim_{h\to 0}\frac{1-9x-9h-1+9x}{h\left(\sqrt{1-9(x+h)}+\sqrt{1-9x}\right)}\\ &= \lim_{h\to 0}\frac{-9h}{h\left(\sqrt{1-9(x+h)}+\sqrt{1-9x}\right)} = \lim_{h\to 0} \frac{-9}{\sqrt{1-9(x+h)}+\sqrt{1-9x}} = \frac{-9}{2\sqrt{1-9x}}\\ \end{align*}

    Be careful when multiplying out the numerator in the first step here. It is easy to lose track of the minus sign (or parentheses for that matter) on the second term.

    The derivative is then $$f'(x) = \frac{-9}{2\sqrt{1-9x}}$$

12. Find the second and third derivatives of the function from Problem 3, $g(x) = x^2.$

    Click for solution

    We found in Problem 3 that the first derivative of $g(x)$ is $g'(x) = 2x.$ The second derivative is then $g''(x) = (g'(x))'.$ To make the notation easier, let's name $g'(x),$ $d(x).$ That is, $d(x) = 2x,$ and we wish to find $d'(x).$

    As always, the first step is to substitute into the definition of the derivative: $$g''(x) = d'(x) = \lim_{h\to 0} \frac{d(x+h) - d(x)}{h} = \lim_{h\to 0}\frac{2(x+h)-2x}{h}$$

    All we need now is a little algebra and we're done:

    $$g''(x) = d'(x) = \lim_{h\to 0} \frac{2x+2h-2x}{h} = \lim_{h\to 0}\frac{2h}{h} = \lim_{h\to 0} (2) = 2$$

    The second derivative of $g$ is then $$g''(x) = 2$.$

    Now we need to find the derivative of this function. Again, let's simplify the notation by renaming $g''(x)$ as $f(x).$ Then $f(x) = 2$ and $g'''(x) = (g''(x))'=f'(x).$

    As with the other problems, we substitute into the definition of the derivative, then do some algebra. In this case, the algebra is not too bad: $$\lim_{h\to 0} \frac{f(x+h)-f(x)}{h} = \lim_{h\to 0}\frac{2 - 2}{h} = \lim_{h\to 0}\frac0h = \lim_{h\to 0} (0) = 0$.$

    The derivative is then $$g'''(x) = f'(x) = 0$$

13. Find the second derivative of the function from Problem 6, $f(x) = 2x^3 - 1.$

    Click for solution

    In Problem 6, we found that the first derivative is $$f'(x) = 6x^2$$ Now we need to find the derivative of this function, $f''(x) = (f'(x))'.$ To make the notation easier, let's rename $f'(x)$ as $d(x)$ and find $f''(x) = d'(x).$

    We first substitute into the definition of the derivative: $$f''(x) = d'(x) = \lim_{h\to 0}\frac{d(x+h)-d(x)}{h} = \lim_{h\to 0}\frac{6(x+h)^2 - 6x^2}{h}$$

    Be sure you evaluate $d(x+h)$ correctly. This is a place where errors often occur.

    Now all we need to do is some algebra and we're done: \begin{align*} f''(x) = d'(x) &= \lim_{h\to 0} \frac{6(x^2 + 2xh +h^2) - 6x^2}{h} = \lim_{h\to 0}\frac{6x^2 + 12xh + 6h^2 - 6x^2}{h}\\ &= \lim_{h\to 0} \frac{12xh + 6h^2}{h} = \lim_{h\to 0}(12x + 6h) = 12x\\ \end{align*}

    The derivative is then $$f''(x) = 12x$$

Assignment Problems

In problems 1 - 24, use the definition of the derivative to find the derivatives of the functions.

$g(x)=10$


$T(y) = -8$


$f(x)=5x+7$


$Q(t) = 1 - 12t$


$f(z) = z^2 + 3$


$R(w) = w^2-8w+20$


$V(t)=6t-t^2$


$Q(t) = 2t^2-8t+10$


$g(z) = 1+10z-7z^2$


$f(x) = 5x-x^3$


$Y(t) = 2t^3+9t+5$


$Z(x) = 2x^3-x^2-x$


$f(t) = \displaystyle\frac{2}{t-3}$


$g(x) = \displaystyle\frac{x+2}{1-x}$


$Q(t) = \displaystyle\frac{t^2}{t+2}$


$f(w) = \sqrt{w+8}$


$V(t) = \sqrt{14+3t}$


$G(x) = \sqrt{2-5x}$


$Q(t)=\sqrt{1+4t}$


$f(x) = \sqrt{x^2+1}$


$W(t) = \displaystyle\frac{1}{\sqrt{t}}$


$g(x) = \displaystyle\frac{4}{\sqrt{1-x}}$


$f(x) = x+\sqrt{x}$


$f(x) = x+\displaystyle\frac{1}{x}$

Note: The following content is adapted from OpenStax and is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License.
Access for free at https://openstax.org/books/calculus-volume-1/pages/1-introduction.

For the functions in problems 25 - 28, (a) sketch the graph, and (b) use the definition of the derivative to show that the function is not differentiable at $x=1.$

25. $f(x) = \left\{\begin{array}{ll} 2\sqrt{x} & 0\leq x\leq 1\\ 3x-1 & x\gt 1\\ \end{array}\right.$
26. $f(x) = \left\{\begin{array}{ll} 3 & x\lt 1\\ 3x & x\geq 1\\ \end{array}\right.$
27. $f(x) = \left\{\begin{array}{ll} -x^2 + 2 & x\leq 1\\ x & x\gt 1\\ \end{array}\right.$
28. $f(x) = \left\{\begin{array}{ll} 2x & x\leq 1\\ \frac{2}{x} & x\gt 1\\ \end{array}\right.$

For the graphs in problems 29 and 30,

1. determine for which values of $x=a$ the limit $\lim\limits_{x\to a} f(x)$ exists but $f$ is not continuous at $x = a.$
2. determine for which values of $x=a$ the function is continuous but not differentiable at $x=a.$

29. 
30. 
31. Use the graph below to evaluate the following derivatives, if they exist: (a) $f'(-0.5),$ (b) $f'(0),$ (c) $f'(1),$ (d) $f'(2),$ (e) $f'(3).$
    

For the following functions, use $f''(x) = \lim\limits_{h\to 0} \displaystyle\frac{f'(x+h)-f'(x)}{h}$ to find $f''(x).$

32. $f(x) = 2-3x$
33. $f(x) = 4x^2$
34. $f(x) = x + \frac{1}{x}$

Determine whether each of the following statements is true or false. Justify your answer.

35. Every function has a derivative.
36. A continuous function has a continuous derivative.
37. A continuous function has a derivative.
38. If a function is differentiable, it is continuous.