3.3 The derivative

What are the properties of $f'$ when $f$ is a line?

(a), (d)

The function shown is a line, so it has a constant slope–(a) . Since the function is always increasing, $f'$ is always positive, so also (d) holds. Remark: it does not matter that the function itself is sometimes negative; the slope is always positive because the function is always increasing. Also, since the slope is constant, $f'$ is neither increasing nor decreasing: it is the function that is increasing, not its derivative.

Be very careful not to confuse $f$ and $f'$.

(e)

The function is always decreasing, so $f'$ is always negative, option (e). However, the function alternates between being more and less steep, so $f'$ alternates between increasing and decreasing several times, and no other options hold.

Remark: $f$ is always positive, but (d) does not hold!

Be very careful not to confuse $f$ and $f'$.

(b)

At the left end of the graph, $f$ is decreasing rapidly, so $f'$ is a strongly negative number. Then as we move towards $x=0$, $f$ decreases less rapidly, so $f'$ is a less strongly negative number. As we pass 0, $f$ increases, so $f'$ is a positive number. As we move to the right, $f$ increases more and more rapidly, so $f'$ is an increasing positive number. This description tells us that $f'$ increases for the entire range shown. So (b) holds, but not (a) or (c). Since $f'$ is negative to the left of the $y$ axis, and positive to the right of it, also (d) and (e) do not hold.

By definition, $f(x) = x^3$ is differentiable at $x = 0$ if the limit

exists.

By definition, $f(x) = x^3$ is differentiable at $x = 0$ if the limit

exists.

The slope has to look “the same" from the left and the right.

$x=-1$ and $x=3$

$f'(-1)$ does not exist, because to the left of $x=-1$ the slope is a pretty big positive number (looks like around $+1$) and to the right the slope is $-1/4$. Since the derivative involves a limit, that limit needs to match the limit from the left and the limit from the right. The sharp angle made by the graph at $x=-1$ indicates that the left and right limits do not match, so the derivative does not exist.

$f'(3)$ also does not exist. One way to see this is to notice that the function is discontinuous here. More viscerally, note that $f(3)=1$, so as we take secant lines with one endpoint $(3,1)$, and the other endpoint just to the right of $x=3$, we get slopes that are more and more strongly negative, as shown in the picture below. If we take the limit of the slopes of these secant lines as $x$ goes to $3$ from the right, we get $-\infty$. (This certainly doesn't match the slope from the left, which is $-\frac{1}{4}$.)

At $x=-3$, there is some kind of “change" in the graph; however, it is a smooth curve, so the derivative exists here.

Use the definition of the derivative, and what you know about limits.

True. (Contrast to Question 7.)

True. The definition of the derivative tells us that

if it exists. We know from our work with limits that if both one-sided limits
$\ds\lim_{h \to 0^-}\frac{f(a+h)-f(a)}{h}$ and $\ds\lim_{h \to 0^+}\frac{f(a+h)-f(a)}{h}$ exist and are equal to each other, then $\ds \lim_{h \to 0}\dfrac{f(a+h)-f(a)}{h}$ exists and has the same value as the one-sided limits. So, since the one-sided limits exist and are equal to one, we conclude $f'(a)$ exists and is equal to one.

Consider continuity.

In general, false. (Contrast to Question 6.)

In general, this is false. The key problem that can arise is that $f(x)$ might not be continuous at $x=1$. One example is the function

where $f'(x)=1$ whenever $x \neq 0$ (so in particular, $\ds\lim_{x \to 0^-}f'(x)=\ds\lim_{x \to 0^+} f'(x)=1$) but $f'(0)$ does not exist.

There are two ways to see that $f'(0)$ does not exist. One is to notice that $f$ is not continuous at $x=0$.

Another way to see that $f'(0)$ does not exist is to use the definition of the derivative. Remember, in order for a limit to exist, both one-sided limits must exist. Let's consider the limit from the left. If $h \to 0^-$, then $h<0$, so $f(h)$ is equal to $h$ (not $h-1$).

In particular, this limit does not exist. Since the one-sided limit does not exist,

and so $f'(0)$ does not exist.

Look at the definition of the derivative. Your answer will be a fraction.

metres per second

Using the definition of the derivative,

The units of the numerator are meters, and the units of the denominator are seconds (since the denominator comes from the change in the input of the function). So, the units of $s'(t)$ are metres per second.

Remark: we learned that the derivative of a position function gives velocity. In this example, the position is given in metres, and the velocity is measured in metres per second.

You need a point (given), and a slope (derivative).

$y-6=3(x-1)$, or $y=3x +3$

We can use point-slope form to get the equation of the line, if we have a point and its slope. The point is given: $(1,6)$. The slope is the derivative:

So our slope is 3, which gives a line of equation $y-6=3(x-1)$.

You'll need to add some fractions.

$\dfrac{-1}{x^2}$

We set up the definition of the derivative.

You don't have to take the limit from the left and right separately–things will cancel nicely.

By definition

In particular, the limit exists, so the derivative exists (and is equal to zero).

By definition

In particular, the limit exists, so the derivative exists (and is equal to zero).

You might have to add fractions

$\dfrac{-2}{(x+1)^2}$

We set up the definition of the derivative.

$\dfrac{-2x}{[x^2+3]^2}$

Your limit should be easy.

1

The slope of the tangent line is the derivative. We set this up using the same definition of the derivative that we always do. This limit is hard to take for general $x$, but easy when $x=0$.

So, the slope of the tangent line is 1.

add fractions

$f'(x)=-\dfrac{2}{x^3}$

For $f$ to be differentiable at $x=2$, two things must be true: it must be continuous at $x=2$, and the derivative from the right must equal the derivative from the left.

$a=4$, $b=-4$

When $x$ is not equal to 2, then the function is differentiable– the only place we have to worry about is when $x$ is exactly 2.

In order for $f$ to be differentiable at $x=2$, it must also be continuous at $x=2$. This forces $x^2\big|_{x=2}=\big[ax+b\big]_{x=2}$ or

In order for a limit to exist, the left- and right-hand limits must exist and be equal to each other. Since a derivative is a limit, in order for $f$ to be differentiable at $x=2$, the left hand derivative of $ax+b$ at $x=2$ must be the same as the right hand derivative of $x^2$ at $x=2$. Since $ax+b$ is a line, its derivative is $a$ everywhere. We've already seen the derivative of $x^2$ is $2x$, so we need

So, the values of $a$ and $b$ that makes $f$ differentiable everywhere are $a=4$ and $b=-4$.

After you plug in $f(x)$ to the definition of a derivative, you'll want to multiply and divide by the conjugate $\sqrt{1+x+h}+\sqrt{1+x}$.

$f'(x)=\dfrac{1}{2\sqrt{1+x}}$ when $x>-1$; $f'(x)$ does not exist when $x \leq -1$.

We plug in $f(x)$ to the definition of a derivative. To evaluate the limit, we multiply and divide by the conjugate of the numerator, then simplify.

The domain of the function is $[-1,\infty)$. In particular, $f(x)$ is defined when $x=-1$. However, $f'(x)$ is not defined when $x=-1$, so $f'(x)$ only exists over $(-1,\infty)$.

Remark: $\ds\lim_{x \to -1^+}f'(x)=\infty$, so the tangent line to $f(x)$ at the point $x=-1$ has a vertical slope.

When you're finding the derivative, you'll need to cancel a lot on the numerator, which you can do by expanding the polynomials.

$v(t)=4t^3-2t$

Recall the velocity is exactly the derivative.

So, the velocity is given by $v(t)=4t^3-2t$.

You'll need to look at limits from the left and right. The fact that $f(0)=0$ is useful for your computation. Recall that if $x<0$ then $\sqrt{x^2}=|x|=-x$.

No, it does not.

The function is differentiable at $x=0$ if the following limit:

exists (note that we used the fact that $f(0)=0$ as per the definition of the first branch which includes the point $x=0$). We start by computing the left limit. For this computation, recall that if $x<0$ then $\sqrt{x^2}=|x|=-x$.

Now, from the right:

Since the limit from the left does not equal the limit from the right, the derivative does not exist at $x=0$.

You'll need to look at limits from the left and right. The fact that $f(0)=0$ is useful for your computation.

No, it does not.

The function is differentiable at $x=0$ if the following limit:

exists (note that we used the fact that $f(0)=0$ as per the definition of the first branch which includes the point $x=0$).

We start by computing the left limit.

Now, from the right:

Since the limit from the left does not equal the limit from the right, the derivative does not exist at $x=0$.

You'll need to look at limits from the left and right. The fact that $f(0)=0$ is useful for your computation.

Yes, it is.

The function is differentiable at $x=0$ if the following limit:

exists (note that we used the fact that $f(0)=0$ as per the definition of the first branch which includes the point $x=0$). We compute left and right limits; so

and

This last limit equals $0$ (see Question 23 in 2.1.1 for a similar example).

Since the left and right limits match (they're both equal to $0$), we conclude that indeed $f(x)$ is differentiable at $x=0$ (and its derivative at $x=0$ is actually equal to $0$).

You'll need to look at limits from the left and right. The fact that $f(1)=0$ is useful for your computation.

Yes, it is.

The function is differentiable at $x=1$ if the following limit:

exists (note that we used the fact that $f(1)=0$ as per the definition of the first branch which includes the point $x=0$). We compute left and right limits; so

and

For this last limit, note that $\left|\sin\left(\frac{1}{x-1}\right)\right| \le 1$, so $\left|(x-1)\cdot \sin\left(\frac{1}{x-1}\right)\right| \le |x-1|$. That is, the `sine' part of the product can only make the $(x-1)$ part closer to 0, not farther from 0. Since $\lim\limits_{x \to 1^+}(x-1)=0$, then also $\lim\limits_{x\to 1^+}(x-1)\cdot \sin\left(\frac{1}{x-1}\right)=0$.

Since the left and right limits match (they're both equal to $0$), we conclude that indeed $f(x)$ is differentiable at $x=1$ (and its derivative at $x=1$ is actually equal to $0$).

There's lots of room between $0$ and $\frac{1}{8}$; see what you can do with it.

Many answers are possible; here is one.

Many answers are possible; here is one.

The key is to realize that the few points you're given suggest a pattern, but don't guarantee it. You only know nine points; anything can happen in between.

Set up your usual limit, then split it into two pieces

At step ($*$), we use the limit law that $\displaystyle\lim_{x \rightarrow a} \left[F(x)+G(x)\right] = \displaystyle\lim_{x \rightarrow a} F(x)+\displaystyle\lim_{x \rightarrow a}G(x)$, as long as $\displaystyle\lim_{x \rightarrow a} F(x)$ and $\displaystyle\lim_{x \rightarrow a}G(x)$ exist. Because the problem states that $f'(x)$ and $g'(x)$ exist, we know that $\displaystyle\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$ and $\displaystyle\lim_{h \rightarrow 0} \frac{g(x+h)-g(x)}{h}$ exist, so our work is valid.

At step ($*$), we use the limit law that $\displaystyle\lim_{x \rightarrow a} \left[F(x)+G(x)\right] = \displaystyle\lim_{x \rightarrow a} F(x)+\displaystyle\lim_{x \rightarrow a}G(x)$, as long as $\displaystyle\lim_{x \rightarrow a} F(x)$ and $\displaystyle\lim_{x \rightarrow a}G(x)$ exist. Because the problem states that $f'(x)$ and $g'(x)$ exist, we know that $\displaystyle\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$ and $\displaystyle\lim_{h \rightarrow 0} \frac{g(x+h)-g(x)}{h}$ exist, so our work is valid.