4.7 Inverse trig functions and their derivatives

Remember that only certain numbers can come out of sine and cosine, but any numbers can go in.

(a) $(-\infty,\infty)$ (b) all integer multiples of $\pi$ (c) $[-1,1]$

(a) We can plug any number into the cosine function, and it will return a number in $[-1,1]$. The domain of $\arcsin x$ is $[-1,1]$, so any number we plug into cosine will give us a valid number to plug into arcsine. So, the domain of $f(x)$ is all real numbers.

(b) We can plug any number into the cosine function, and it will return a number in $[-1,1]$. The domain of $\arccsc x$ is $(-\infty,-1] \cup [1,\infty)$, so in order to have a valid number to plug into arccosecant, we need $\cos x = \pm 1$. That is, the domain of $g(x)$ is all values $x=n\pi$ for some integer $n$.

(c) The domain of arccosine is $[-1,1]$. The domain of sine is all real numbers, so no matter what number arccosine spits out, we can safely plug it into sine. So, the domain of $h(x)$ is $[-1,1]$.

What is the range of the arccosine function?

False

False: $\cos t=1$ for infinitely many values of $t$; arccosine gives only the single value $t=0$ for which $\cos t=1$ and $0 \leq t \leq \pi$. The particle does not start moving until $t=10$, so $t=0$ is not in the domain of the function describing its motion.

The particle will have height $1$ at time $2\pi n$, for any integer $n \geq 2$.

A one–to–one function passes the horizontal line test. To graph the inverse of a function, reflect it across the line $y=x$.

First, we restrict the domain of $f$ to force it to be one–to–one. There are many intervals we could choose over which $f$ is one–to–one, but the question asks us to contain $x=0$ and be as large as possible; this leaves us with the following restricted function:

The inverse of a function swaps the role of the input and output; so if the graph of $y=f(x)$ contains the point $(a,b)$, then the graph of $Y=f^{-1}(X)$ contains the point $(b,a)$. That is, the graph of $Y=f^{-1}(X)$ is the graph of $y=f(x)$ with the $x$-coordinates and $y$-coordinates swapped. (So, since $y=f(x)$ crosses the $y$-axis at $y=1$, then $Y=f^{-1}(X)$ crosses the $X$-axis at $X=1$.) This swapping is equivalent to reflecting the curve $y=f(x)$ over the line $y=x$.

Remark: while you're getting accustomed to inverse functions, it is sometimes clearer to consider $y=f(x)$ and $Y=f^{-1}(X)$: using slightly different notations for $x$ (the input of $f$, hence the output of $f^{-1}$) and $X$ (the input of $f^{-1}$, which comes from the output of $f$). However, the convention is to use $x$ for the inputs of both functions, and $y$ as the outputs of both functions, as is written on the graph above.

Your answer will depend on $a$. The arcsine function alone won't give you every value.

The tangent line is horizontal when $0=y'=a-\sin x$. That is, when $a=\sin x$.

Remark: when $a=1$, then

Similarly, when $a=-1$,

So, if we try to use the descriptions in the third bullet point to describe points where the tangent line is horizontal when $|a|=1$, we get the correct points but each point is listed twice. This is why we separated the case $|a|=1$ from the case $|a|<1$.

In order for $x$ to be in the domain of $f$, you must be able to plug $x$ into both arcsine and arccosecant.

Domain: $x=\pm 1$. Not differentiable anywhere.

The function $\arcsin x$ is only defined for $|x| \leq 1$, and the function $\arccsc x$ is only defined for $|x| \geq 1$, so $f(x)$ has domain $|x|=1$. That is, $x=\pm1$.

In order for $f(x)$ to be differentiable at a point, it must exist in an open interval around that point. (See Definition 3.3.3.) Since our function does not exist over any open interval, $f(x)$ is not differentiable anywhere.

So, actually, $f(x)$ is a pretty boring function, which we can entirely describe as: $f(-1)=-\pi$ and $f(1)=\pi$.

For the domain of $f$, remember the domain of arcsine is $[-1,1]$.

$f'(x)=\dfrac{1}{\sqrt{9-x^2}}$; domain of $f$ is $[-3,3]$.

Using the chain rule,

Since the domain of arcsine is $[-1,1]$, and we are plugging in $\dfrac{x}{3}$ to arcsine, the values of $x$ that we can plug in are those that satisfy $-1 \le \dfrac{x}{3} \leq 1$, or $-3\leq x \leq 3$. So the domain of $f$ is $[-3,3]$.

The domain of $\arccos(t)$ is $[-1,1]$, but you also have to make sure you aren't dividing by zero.

$f'(t)=\dfrac{-\frac{t^2-1}{\sqrt{1-t^2}}-2t\arccos t}{(t^2-1)^2}$, and the domain of $f(t)$ is $(-1,1)$.

Using the quotient rule,

The domain of arccosine is $[-1,1]$, and since $t^2-1$ is in the denominator, the domain of $f$ requires $t^2-1 \neq 0$, that is, $t \neq \pm 1$. So the domain of $f(t)$ is $(-1,1)$.

$\ds\diff{}{x}\left\{\arcsec x\right\} = \dfrac{1}{|x|\sqrt{x^2-1}}$, and the domain of $\arcsec x$ is $|x|\ge1$.

The domain of $f(x)$ is all real numbers, and $f'(x)=\dfrac{-2x}{(x^2+2)\sqrt{x^4+4x^2+3}}$.

The domain of $\arcsec x$ is $|x| \geq 1$: that is, we can plug into arcsecant only values with absolute value greater than or equal to one. Since $-x^2-2 \leq -2$, every real value of $x$ gives us an acceptable value to plug into arcsecant. So, the domain of $f(x)$ is all real numbers.

To differentiate, we use the chain rule. Remember $\ds\diff{}{x}\left\{\arcsec x\right\} = \dfrac{1}{|x|\sqrt{x^2-1}}$.

The domain of $\arctan(x)$ is all real numbers.

$f'(x)=\dfrac{1}{a^2+x^2}$ and the domain of $f(x)$ is all real numbers.

We use the chain rule, remembering that $a$ is a constant.

The domain of arctangent is all real numbers, so the domain of $f(x)$ is also all real numbers.

The domain of $\arcsin x$ is $[-1,1]$, and the domain of $\sqrt{x}$ is $x \geq 0$.

$f'(x)=\arcsin x$, and the domain of $f(x)$ is $[-1,1]$.

We differentiate using the product and chain rules.

The domain of $\arcsin x$ is $[-1,1]$, and the domain of $\sqrt{1-x^2}$ is all values of $x$ so that $1-x^2 \geq 0$, so $x$ in $[-1,1]$. Therefore, the domain of $f(x)$ is $[-1,1]$.

This occurs only once.

$x=0$

We differentiate using the chain rule:

This is zero exactly when $x=0$.

The answer is a very simple expression.

$\ds\diff{}{x}\{\arcsin x + \arccos x\}=0$

Using formulas you should memorize from this section,

Remark: the only functions with derivative equal to zero everywhere are constant functions, so $\arcsin x + \arccos x$ should be a constant. Since $\sin \theta = \cos \left(\frac{\pi}{2}-\theta\right)$, we can set

where $x$ and $\theta$ are the same in both expressions, and $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$. Then

We note here that arcsine is the inverse of the sine function * restricted to* $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. So, since we restricted $\theta$ to this domain, $\sin \theta=x$ really does imply $\arcsin x = \theta$. (For an example of why this matters, note $\sin(2\pi)=0$, but $\arcsin (0)=0 \neq 2\pi$.) Similarly, arccosine is the inverse of the cosine function restricted to $[0,\pi]$. Since $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$, then $0 \leq (\frac{\pi}{2}-\theta) \leq \pi$, so $\cos\left(\frac{\pi}{2}- \theta\right) =x$ really does imply $\arccos x=\frac{\pi}{2}-\theta$.

So,

which means the derivative we were calculating was actually just $\ds\diff{}{x}\left\{\dfrac{\pi}{2}\right\}=0$.

chain rule

$y'=\dfrac{-1}{x^2\sqrt{1-\frac{1}{x^2}}}$

Using the chain rule,

Recall $\ds\diff{}{x}\{\arctan x\}=\dfrac{1}{1+x^2}=(1+x^2)^{-1}$.

$\ds\ddiff{2}{}{x}\{\arctan x\}=\frac{-2x}{(1+x^2)^2}$

$y'=\dfrac{-1}{1+x^2}$

Using the chain rule,

$2x\arctan x+1$

Using the product rule:

You can simplify the expression before you differentiate to remove the trigonometric functions. If $\arctan x =\theta$, then fill in the sides of the triangle below using the definition of arctangent and the Pythagorean theorem:

With the sides labeled, you can figure out $\sin\left(\arctan x\right)=\sin\left(\theta\right)$.

Let $\theta = \arctan x$. Then $\theta$ is the angle of a right triangle that gives $\tan \theta = x$. In particular, the ratio of the opposite side to the adjacent side is $x$. So, we have a triangle that looks like this:

where the length of the hypotenuse came from the Pythagorean Theorem. Now,

From here, we differentiate using the quotient rule:

Let $\theta = \arctan x$. Then $\theta$ is the angle of a right triangle that gives $\tan \theta = x$. In particular, the ratio of the opposite side to the adjacent side is $x$. So, we have a triangle that looks like this:

where the length of the hypotenuse came from the Pythagorean Theorem. Now,

From here, we differentiate using the quotient rule:

Remark: another strategy is to differentiate first, using the chain rule, then draw a triangle to simplify the resulting expression $\ds\diff{}{x}\left\{\sin\left(\arctan x\right)\right\}=\dfrac{\cos(\arctan x)}{1+x^2}$.

You can simplify the expression before you differentiate to remove the trigonometric functions. If $\arcsin x =\theta$, then fill in the sides of the triangle below using the definition of arctangent and the Pythagorean theorem:

With the sides labeled, you can figure out $\cot\left(\arcsin x\right)=\cot\left(\theta\right)$.

Let $\theta = \arcsin x$. Then $\theta$ is the angle of a right triangle that gives $\sin \theta = x$. In particular, the ratio of the opposite side to the hypotenuse is $x$. So, we have a triangle that looks like this:

where the length of the adjacent side came from the Pythagorean Theorem. Now,

From here, we differentiate using the quotient rule:

Let $\theta = \arcsin x$. Then $\theta$ is the angle of a right triangle that gives $\sin \theta = x$. In particular, the ratio of the opposite side to the hypotenuse is $x$. So, we have a triangle that looks like this:

where the length of the adjacent side came from the Pythagorean Theorem. Now,

From here, we differentiate using the quotient rule:

Remark: another strategy is to differentiate first, using the chain rule, then draw a triangle to simplify the resulting expression $\ds\diff{}{x}\left\{\cot\left(\arcsin x\right)\right\}=\frac{-\csc^2(\arcsin x)}{\sqrt{1-x^2}}$.

What is the slope of the line $y=2x+9$?

$(x,y)=\pm\big(\frac{\sqrt{3}}{2},\frac{\pi}{3}\big)$

The line $y=2x+9$ has slope $2$, so we must find all values of $x$ between $-1$ and $1$ ($\arcsin x$ is only defined for these values of $x$) for which $\diff{}{x}\{\arcsin x\}=2$. Evaluating the derivative:

Differentiate using the chain rule.

$x=\dfrac{(2n+1)\pi}{2}$ for any integer $n$

We differentiate using the chain rule:

So if $f'(x)=0$, then $\cos x=0$, and this happens when $x=\dfrac{(2n+1)\pi}{2}$ for any integer $n$. We should check that these points are in the domain of $f$. Arctangent is defined for all real numbers, so we only need to check the domain of cosecant; when $x=\dfrac{(2n+1)\pi}{2}$, then $\sin x=\pm1 \neq 0$, so $\csc x = \dfrac{1}{\sin x}$ exists.

If $g(y)=f^{-1}(y)$, then $f(g(y))=f\left(f^{-1}(y)\right)=y$. Differentiate this last equality using the chain rule.

$g'(y)=\dfrac{1}{1-\sin g(y)}$

Since $g(y)=f^{-1}(y)$,

Now, we can differentiate with respect to $y$ using the chain rule.

To simplify notation, let $g(y)=f^{-1}(y)$. Simplify and differentiate $g(f(x))$.

$\dfrac{1}{2}$

Write $g(y)=f^{-1}(y)$. Then $g(f(x))=x$, so differentiating both sides (using the chain rule), we see

What we want is $g'(\pi-1)$, so we need to figure out which value of $x$ gives $f(x)=\pi-1$. A little trial and error leads us to $x=\frac{\pi}{2}$.

Since $f'(x)=2-\cos(x)$, $f'\left(\frac{\pi}{2}\right)=2-0=2$:

To simplify notation, let $g(y)=f^{-1}(y)$. Simplify and differentiate $g(f(x))$.

$\dfrac{1}{e+1}$

Write $g(y)=f^{-1}(y)$. Then $g(f(x))=x$, so differentiating both sides (using the chain rule), we see

What we want is $g'(e+1)$, so we need to figure out which value of $x$ gives $f(x)=e+1$. A little trial and error leads us to $x=1$.

It remains only to note that $f'(x)=e^x+1$, so $f'(1)=e+1$