4.2 Trigonometric functions and their derivatives

A horizontal tangent line is where the graph appears to “level off."

The graph $f(x)=\sin x$ has horizontal tangent lines precisely at those points where $\cos x=0$.

The graph $f(x)=\sin x$ has horizontal tangent lines precisely at those points where $\cos x=0$. This must be true, since $\diff{}{x}\{\sin x\}=\cos x$: where the derivative of sine is zero, cosine itself is zero.

You are going to mark there points on the sine graph where the graph is the steepest, going up.

The graph $f(x)=\sin x$ has maximum slope at those points where $\cos x$ has a maximum. That is, where $\cos x = 1$.

The graph $f(x)=\sin x$ has maximum slope at those points where $\cos x$ has a maximum. This makes sense, because $f'(x)=\cos x$: the maximum values of the slope of sine correspond to the maximum values of cosine.

$h'(t)$ gives the velocity of the particle, and $h''(t)$ gives its acceleration–the rate the velocity is changing.

speeding up

The velocity of the particle is given by $h'(t)=\sin t$. Note $0<1<\pi$, so $h'(1)>0$–the particle is rising (moving in the positive direction, in this case “up"). The acceleration of the particle is $h''(t)=\cos t$. Since $0<1<\frac{\pi}{2}$, $h''(t)>0$, so $h'(t)$ is increasing: the particle is moving up, and it's doing so at an increasing rate. So, the particle is speeding up.

$h'(t)$ gives the velocity of the particle, and $h''(t)$ gives its acceleration–the rate the velocity is changing. Be wary of signs–as in legends, they may be misleading.

slower

For this problem, remember that velocity has a sign indicating direction, while speed does not.

The velocity of the particle is given by $h'(t)=3t^2-2t-5$. At $t=1$, the velocity of the particle is $-4$, so the particle is moving downwards with a speed of 4 units per second. The acceleration of the particle is $h''(t)=6t-2$, so when $t=1$, the acceleration is (positive) $4$ units per second per second. That means the velocity (currently $-4$ units per second) is becoming a bigger number–since the velocity is negative, a bigger number is closer to zero, so the speed of the particle is getting smaller. (For instance, a velocity of $-3$ represents a slower motion than a velocity of $-4$.) So, the particle is slowing down at $t=1$.

To show that two functions are unequal, you can show that one input results in different outputs.

(a) true (b) true (c) false

For (a) and (b), notice the following:

The fourth derivative is $\sin x$ is $\sin x$, and the fourth derivative of $\cos x$ is $\cos x$, so (a) and (b) are true.

So, $\ds\ddiff{4}{}{x} \tan x =8\sec^2x\tan^3x+16\sec^4x\tan x$. It certainly seems like this is not the same as $\tan x$, but remember that sometimes trig identities can fool you: $\tan^2x+1=\sec^2x$, and so on. So, to be absolutely sure that these are not equal, we need to find a value of $x$ so that the output of one is not the same as the output of the other. When $x=\frac{\pi}{4}$:

So, (c) is false.

You need to memorize the derivatives of sine, cosine, and tangent.

$f'(x)=\cos x - \sin x + \sec^2 x$

You should memorize the derivatives of sine, cosine, and tangent.
$f'(x)=\cos x - \sin x + \sec^2 x$

There are infinitely many values. You need to describe them all.

$x=\frac{\pi}{4}+\pi n$, for any integer $n$.

$f'(x)=\cos x - \sin x$, so $f'(x)=0$ precisely when $\sin x = \cos x$. This happens at $\pi/4$, but it also happens at $5\pi/4$. By looking at the unit circle, it is clear that $\sin x = \cos x$ whenever $x = \frac{\pi}{4}+\pi n$ for some integer $n$.

Simplify first.

0

The identity won't help you.

$f'(x)=2(\cos^2 x - \sin ^2 x)$

It is true that $2\sin x \cos x = \sin (2x)$, but we don't know the derivative of $\sin(2x)$. So, we use the product rule:

$f'(x)=e^x(\cot x - \csc^2 x)$

Quotient rule

$f'(x)=\dfrac{2+3 \sec x + 2 \sin x -2\tan x \sec x+3\sin x \tan x }{(\cos x + \tan x)^2}$

We use the quotient rule.

$f'(x)=\dfrac{5\sec x \tan x - 5 \sec x - 1}{e^x}$

We use the quotient rule.

$f'(x)=(e^x+\cot x)(30x^5+\csc x \cot x)+(e^x-\csc^2x)(5x^6-\csc x)$

We use the product rule:

Use an identity.

$-\sin(\theta)$

We don't know how to differentiate this function as it is written, but an identity helps us. Since $\sin\left(\frac{\pi}{2}-\theta \right)=\cos \theta$, we see $f'(\theta)=\diff{}{\theta}\{\cos \theta\}=-\sin(\theta)$.

How can you move the negative signs to a location that you can more easily deal with?

$f'(x)=-\cos x - \sin x$

We know the derivative of $\sin x$, but not of $\sin(-x)$. So we re-write $f(x)$ using identities:

Apply the quotient rule.

$\left(\dfrac{\cos\theta+\sin\theta}{\cos\theta-\sin\theta}\right)^2+1$

We apply the quotient rule.

The only spot to worry about is when $x=0$. For $f(x)$ to be differentiable, it must be continuous, so first find the value of $b$ that makes $f$ continuous at $x=0$. Then, find the value of $a$ that makes the derivatives from the left and right of $x=0$ equal to each other.

$a=0$, $b=1$.

In order for $f$ to be differentiable at $x=0$, it must also be continuous at $x=0$. This forces

or $b=1$. In order for $f$ to be differentiable at $x=0$, we need the limit

to exist. This is the case if and only if the two one–sided limits

and

exist and are equal. Because $\cos(x)$ is differentiable at $x=0$ we have

So, we need $a=0$ and $b=1$.

$y - \pi = 1\cdot (x-\pi/2)$

We compute the derivative of $\cos(x)+2x$ as being $-\sin(x)+2$, which evaluated at $x=\frac{\pi}{2}$ yields $-1+2=1$. Since we also compute $\cos(\pi/2)+2(\pi/2)=0+\pi$, then the equation of the tangent line is

Compare this to one of the forms given in the text for the definition of the derivative.

$-\sin(2015)$

This limit represents the derivative computed at $x=2015$ of the function $f(x)=\cos(x)$. To see this, simply use the definition of the derivative at $a=2015$ with $f(x)=\cos x$:

Since the derivative of $f(x)$ is $-\sin(x)$, its value at $x=2015$ is exactly $-\sin(2015)$.

Compare this to one of the forms given in the text for the definition of the derivative.

$-\sqrt{3}/2$

This limit represents the derivative computed at $x=\pi/3$ of the function $f(x)=\cos x$. To see this, simply use the definition of the derivative at $a=\pi/3$ with $f(x)=\cos x$:

Since the derivative of $f(x)$ is $-\sin x$, then its value at $x=\pi/3$ is exactly
$-\sin(\pi/3)=-\sqrt{3}/2$.

Compare this to one of the forms given in the text for the definition of the derivative.

$-1$

This limit represents the derivative computed at $x=\pi$ of the function $f(x)=\sin(x)$. To see this, simply use the definition of the derivative at $a=\pi$ with $f(x)=\sin x$:

Since the derivative of $f(x)$ is $\cos(x)$, then its value at $x=\pi$ is exactly
$\cos(\pi)=-1$.

$\tan \theta = \dfrac{\sin \theta}{\cos \theta}$

So, using the quotient rule,

So, using the quotient rule,

In order for a derivative to exist, the function must be continuous, and the derivative from the left must equal the derivative from the right.

$a=-\frac{2}{3}$, $b=2$

In order for the function $f(x)$ to be continuous at $x=0$, the left half formula $ax+b$ and the right half formula $\dfrac{6\cos x}{2+\sin x+\cos x}$ must match up at $x=0$. This forces

In order for the derivative $f'(x)$ to exist at $x=0$, the limit $\ds\lim_{h \rightarrow 0} \dfrac{f(h)-f(0)}{h}$ must exist. In particular, the limits $\ds\lim_{h \rightarrow 0^-} \dfrac{f(h)-f(0)}{h}$ and $\ds\lim_{h \rightarrow 0^+} \dfrac{f(h)-f(0)}{h}$ must exist and be equal to each other.

When $h \to 0^-$, this means $h<0$, so $f(h)=ah+b=ah+2$. So:

Similarly, when $h \to 0^+$, then $h>0$, so $f(h)=\dfrac{6\cos h}{1+\sin h + \cos h}$ and

Since the limits from the left and right must be equal, this forces

There are infinitely many places where it does not exist.

All values of $x$ except $x=\frac{\pi}{2}+n\pi$, for any integer $n$.

In order for $f'(x)$ to exist, $f(x)$ has to exist. We already know that $\tan x$ does not exist whenever $x=\frac{\pi}{2}+n\pi$ for any integer $n$. If we look a little deeper, since $\tan x = \frac{\sin x}{\cos x}$, the points where tangent does not exist correspond exactly to the points where cosine is zero.

From its graph, tangent looks like a smooth curve over its domain, so we might guess that everywhere tangent is defined, its derivative is defined. We can check this: $f'(x) = \sec^2 x = \left(\frac{1}{\cos x}\right)^2$. Indeed, wherever $\cos x$ is nonzero, $f'$ exists.

So, $f'(x)$ exists for all values of $x$ except when $x=\frac{\pi}{2}+n\pi$ for some integer $n$.

The function is differentiable whenever $x^2+x-6\ne 0$ since the derivative equals

which is well-defined unless $x^2+x-6=0$. We solve $x^2+x-6=(x-2)(x+3)=0,$ and get $x=2$ and $x=-3$. So, the function is differentiable for all real values $x$ except for $x=2$ and for $x=-3$.

The function is differentiable whenever $x^2+x-6\ne 0$ since the derivative equals

which is well-defined unless $x^2+x-6=0$. We solve $x^2+x-6=(x-2)(x+3)=0,$ and get $x=2$ and $x=-3$. So, the function is differentiable for all real values $x$ except for $x=2$ and for $x=-3$.

The function is differentiable whenever $\sin(x)\ne 0$ since the derivative equals

which is well-defined unless $\sin x = 0$. This happens when $x$ is an integer multiple of $\pi$. So, the function is differentiable for all real values $x$ except $x=n\pi,$, where $n$ is any integer.

The function is differentiable whenever $\sin(x)\ne 0$ since the derivative equals

which is well-defined unless $\sin x = 0$. This happens when $x$ is an integer multiple of $\pi$. So, the function is differentiable for all real values $x$ except $x=n\pi,$, where $n$ is any integer.

$y - 1 = 2\cdot (x-\pi/4)$

We compute the derivative of $\tan(x)$ as being $\sec^2(x)$, which evaluated at $x=\frac{\pi}{4}$ yields $2$. Since we also compute $\tan(\pi/4)=1$, then the equation of the tangent line is

$y=2x+2$

We compute the derivative $y' = \cos(x)-\sin(x)+e^x$, which evaluated at $x=0$ yields $1-0+1 = 2$. Since we also compute $y(0)=0+1+1=2$, the equation of the tangent line is

ie $y=2x+2$.

$x = \frac{3\pi}{4}+n\pi$ for any integer $n$.

We are asked to solve $f'(x)=0$. That is, $e^x[\sin x + \cos x]=0$. Since $e^x$ is always positive, that means we need to find all points where $\sin x + \cos x =0$. That is, we need to find all values of $x$ where $\sin x = - \cos x$. Looking at the unit circle, we see this happens whenever $x = \frac{3\pi}{4}+n\pi$ for any integer $n$.

Recall $|x|=\left{\begin{array}{rl} x&x\ge 0\ -x&x<0 \end{array}\right.$. To determine whether$h(x)$is differentiable at$x=0$, use the definition of the derivative.

$h'(x)=\left{\begin{array}{rl} \cos x&x> 0\ -\cos x&x<0 \end{array}\right.$ It exists for all $x \neq 0$.

As usual, when dealing with the absolute value function, we can make things a little clearer by splitting it up into two pieces.

So,

where we used the identity $\sin(-x)=-\sin x$. From here, it's easy to see $h'(x)$ when $x$ is anything other than zero.