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4.6: Transcendental Functions

Difficulty Level: At Grade Created by: CK-12
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Inverse Functions


The idea behind inverse functions is not difficult. Recall that for a function to be well-defined each \begin{align*}x\end{align*} must correspond to one and only one \begin{align*}y\end{align*}. If an inverse function is to exist, then each \begin{align*}y\end{align*} must correspond to one and only one \begin{align*} x\end{align*}. Geometrically this means that any horizontal line (which is the function \begin{align*}y=\;\mathrm{constant}\end{align*}) must touch the graph at only one point. Otherwise the \begin{align*}y\end{align*}-value corresponding to this horizontal line would have be the image of more than one \begin{align*}x\end{align*}.

We see here that the function \begin{align*}f(x)\end{align*} has no inverse since the horizontal like \begin{align*}y=-3\end{align*} intersects the plot at three different points. Therefore there are \begin{align*}3\end{align*} different values of \begin{align*}x\end{align*} for which \begin{align*}f(x)=-3\end{align*} so that \begin{align*}f^{-1} (-3)\end{align*} is not well-defined.

Geometrically, the derivative of a function is the slope of its tangent line. Since the inverse of a function is just its reflection across the line \begin{align*}y=x\end{align*} we see that the tangent line at a point \begin{align*}(x,f(x))\end{align*} will just be reflected into its reciprocal. Therefore, wherever \begin{align*}\frac{dy}{dx}\end{align*} exists and is nonzero, its reciprocal exists and so the inverse is differentiable. This can be interpreted by saying that the function nearly fails to be invertible by becoming horizontal somewhere. That is, a function only fails to be invertible if it turns around somewhere. Otherwise, the function will not fail the horizontal line test as it would be increasing or decreasing everywhere. To fail the horizontal line test it must go from increasing to decreasing or vice-versa. Now, we saw earlier that the function only changes direction in a differentiable way if its derivative is zero when it changes direction.


To teach this material it is valuable to have students think about the material geometrically. For this reason, any question should first ask students to graph the function in question and then use this to understand why and how it fails or succeeds at being invertible. Similarly, the plot will show them how a function’s inverse may fail to be differentiable. In this context, it should also be pointed out that while a function itself may not be differentiable at a certain point, its inverse may be. For example, the function \begin{align*}f(x)=x^{\frac{1}{3}}\end{align*} is not differentiable at \begin{align*}x=0\end{align*}; however, its inverse \begin{align*}f(x)=x^3\end{align*} is differentiable everywhere:

Exponential and Logarithmic Functions


Exponential and logarithmic functions have huge roles throughout math, science, and business. In the same way that the number \begin{align*}\pi\end{align*} arises naturally from considerations of basic geometry, the exponential and logarithmic functions arise naturally in the context of calculus. When we consider taking how fast a function is changing, i.e. its derivative, and look for a function whose value at each point is equal to this rate, we arrive at the exponential.


A nice idea to have students experiment with these functions is to split them into an even number of small groups or pairs and challenge each group to completely analyze some function like \begin{align*}f(x)=e^{\frac{1}{10} x}\end{align*} or \begin{align*}g(x) = \frac{1}{10} \;\mathrm{Log} (x)\end{align*}. They should plot this, find its derivatives and plot these, and find the inverse function. Then, each group can be told that some other group has the inverse of their function and they must find out who it is.


To test this material, students can be given problems of exponential growth and logarithmic plots. For example, there are endless problems having to do with interest rates:

1. Suppose you want to invest \begin{align*}\$100\;\mathrm{dollars}\end{align*} into a bank account and leave it for \begin{align*}10\;\mathrm{years}\end{align*} in an account that has an annual rate of \begin{align*}r=5\%\end{align*}. a. If the interest is compounded yearly, how much will it be worth in \begin{align*}10\;\mathrm{years}\end{align*}?

i. ANSWER: \begin{align*}\$100(1.05)^{10}=\$162.89\end{align*}

b. If the interest is compounded monthly, how much will there be in \begin{align*}10\;\mathrm{years}\end{align*}?

i. ANSWER: Interest each month is \begin{align*}\frac{5}{12} \%\end{align*} so total is:

\begin{align*}\$100 \left ( 1+\frac{5}{12 \ 100} \right )^{120} = \$100 \left (\frac{241}{240} \right )^{120} = \$164.70\end{align*}

c. If the interest is compounded daily, how much will you have in \begin{align*}10\;\mathrm{years}\end{align*}?

i. ANSWER: Interest each day will be \begin{align*}\frac{5}{365} \%\end{align*} so the total is:

\begin{align*}\$100 \left ( 1+\frac{5}{365 \ 100} \right )^{3650} = \$100 \left (\frac{7301}{7300} \right )^{3650} = \$164.866\end{align*}

d. If the interest is compounded continuously, how much will it be worth in \begin{align*}10\;\mathrm{years}\end{align*}? i. ANSWER: Performing the same calculation in the limit as \begin{align*} n \to \infty\end{align*} we obtain

\begin{align*}\$100 \left [\lim_{n \to \infty} \left ( 1+\frac{5}{n \ 100} \right )^{10 \ n} \right ]\end{align*}

So that if we then \begin{align*}m=10n\end{align*} we have that:

\begin{align*}\$100 \left [\lim_{n \to \infty} \left ( 1+\frac{\frac{5}{10}}{m} \right )^m \right ] = \$100 \ e^{\frac{5}{10}} = \$ 164.872\end{align*}

2. Below is a plot of the Earth’s population since the year \begin{align*}1750\end{align*}.

a. Draw a plot of this on a logarithmic scale. In other words, this is a plot of population \begin{align*}P\end{align*} vs. time \begin{align*}t\end{align*}. Draw a plot of Log-Population \begin{align*}\;\mathrm{Log}(P)\end{align*} vs. time \begin{align*}t\end{align*}. The approximate linearity of this plot should demonstrates the relationship that:

\begin{align*}\text{Log}(P)=m \ t+b\end{align*}

which implies that:

\begin{align*}P=P_0 \ e^{mt} \qquad \text{where} \qquad P_0 =e^b\end{align*}

b. Find the appropriate constants \begin{align*}P_0\end{align*} and \begin{align*}m\end{align*} so that this relationship holds. This can be done by examining the linear plot to find the \begin{align*}y\end{align*}-intercept \begin{align*}b\end{align*} and the slope \begin{align*}m\end{align*}.

c. Find the average time for the population to double in terms of the constant \begin{align*}m\end{align*} above.

Differentiation and Integration of Logarithmic and Exponential Functions


In general it tends to be difficult for students to simply memorize all of these cumbersome formulas for derivatives. The easiest thing is to teach the simplest formulas:

\begin{align*}\frac{d}{dx} e^x & = e^x \qquad \qquad \qquad (1) \\ \frac{d}{dx} \text{ln} \ (x) & = \frac{1}{x} \qquad \qquad \qquad (2)\end{align*} All other situations can easily be reduced to something like this. As an example, consider the complicated formula for an “exponential function” with arbitrary base \begin{align*}b\end{align*}:


This can be written as:

\begin{align*}b^x = e^{\text{ln}(b^x)} =e^{x \ \text{ln}(b)}\end{align*}

and so we can just use the Chain Rule:

\begin{align*}\frac{d}{dx} b^x = \frac{d}{dx} e^{x \ \text{ln} (b)} =e^{x \ \text{ln}(b)} \frac{d}{dx} (x \ \text{ln} (b) ) = b^x \ \text{ln}(b)\end{align*}

Similarly, the complicated formula for a “logarithmic function” with arbitrary base \begin{align*}b\end{align*}:

\begin{align*}log_b \ x\end{align*}

can be written as:

\begin{align*}log_b \ x = \frac{1}{\text{ln}(b)} \text{ln}(x)\end{align*}

and so we just get:

\begin{align*}\frac{d}{dx} \text{log}_b \ x = \frac{1}{\text{ln}(b)} \frac{d}{dx} \text{ln}(x) = \frac{1}{x \ \text{ln} (b)}\end{align*}


It is best to teach this material by having students practice with transforming the complicated functions into simpler ones. They should certainly have the basic formulas \begin{align*}(1)\end{align*} and \begin{align*}(2)\end{align*} above memorized. But instead of just mindlessly applying the more complicated versions they should understand how to reduce everything to these formulas.  

Exponential Growth and Decay


Reproduction is common to all forms of life, describing the process by which organisms essentially create replicas of themselves. Anytime the elements of a set replicate themselves over cycles, the size of the set grows faster the bigger it is. That is to say, the amount the set grows each cycle is proportional to its current size. This is the hallmark of exponential growth, since an exponential function \begin{align*}f(x)=f(0) \ e^{\;\mathrm{rt}}\end{align*} has a derivative \begin{align*}f'(x)=r \ [f(0) e^{\;\mathrm{rt}}] = r \ f(x)\end{align*} that is proportional to itself.


A nice problem to walk through with the students is the following:

A “forest” is planted with one tree in it, and this tree can spawn \begin{align*}10\end{align*} trees per year. Then if the next year there are \begin{align*}11\end{align*} trees in the forest, the population can grow at a rate of \begin{align*}110\end{align*} trees per year. The next year if there are \begin{align*}121\end{align*} trees, then the population can grow at a rate of \begin{align*}1210\end{align*} trees per year. So the population growth rate \begin{align*}p'(t)=10 \ p(t)\end{align*} where \begin{align*}p(t)\end{align*} is the population at time \begin{align*}t\end{align*}. This equation is written more simply as:

\begin{align*}p' = 10p\end{align*}

and is a standard differential equation. The initial population is given as \begin{align*}p(0)=1\end{align*} trees, then we can solve this by thinking about what function gives itself times a constant back. Well the function \begin{align*}e^t\end{align*} gives itself back and we can modify it slightly to see that:

\begin{align*}\frac{d}{dt}(e^{10t}) = 10 \ (e^{10t})\end{align*}

However, this is true for any multiplicative constant in front since for any \begin{align*}C\end{align*} whatsoever we have:

\begin{align*}\frac{d}{dt}(Ce^{10t}) =10 \ (Ce^{10t})\end{align*}

so it would seem that our differential equation is solved by the function:

\begin{align*}p = Ce^{10t}\end{align*}

and we can use the fact that \begin{align*}p(0)=1\end{align*} to find \begin{align*}C\end{align*}:

\begin{align*}\text{at} \ t=0: \qquad 1=Ce^{(10)(0)} =C\end{align*}

so \begin{align*}C=1\end{align*} and the population after \begin{align*}t\end{align*} years is given by:


After completing this, it is worthwhile to erase the board and have students attempt to work either alone or in groups to solve a similar problem. This will force them to try recalling each step along the way.  

Derivatives and Integrals Involving Inverse Trigonometric Functions


A quick glance at the plot of any trigonometric function:

shows that they all fail the horizontal line test miserably. However we can restrict their domains so that over these new functions defined only on the restricted domains do have inverses. Their plots over these restricted domains look like:

These functions, restricted to the smaller domains, clearly have no problems passing the horizontal-line test. Therefore on these domains the functions are invertible and the inverses are determined quite easily.


In order to teach this material in the most accessible way possible the means of obtaining the formulae presented should be shown. In fact, this is not difficult by any means. For example, consider finding the formula for \begin{align*}\frac{d}{dx} \mathrm{Sin} ^{-1} (x)\end{align*}:

Let: \begin{align*}u = \mathrm{Sin}(x)\end{align*}

Then: \begin{align*}x = \mathrm{Sin} ^{-1} (u)\end{align*}

And so:\begin{align*} \frac{dx}{du} = \frac{d}{du} \mathrm{Sin} ^{-1} (u) \end{align*} But also: \begin{align*}\frac{dx}{du} = \frac{1}{\frac{du}{dx}} = \frac{1}{\mathrm{Cos}(x)} = \frac{1}{\sqrt{1-\mathrm{Sin} ^2 (x)}} = \frac{1}{\sqrt{1-u^2}}\end{align*}

So we conclude that:\begin{align*} \frac{d}{du} \mathrm{Sin} ^{-1} (u) = \frac{1}{\sqrt{1-u^2}}\end{align*}

The other formulae can be similarly derived and leaving this out will only make the material seem more odd and difficult to swallow. Students will have an easier time using, manipulating, and recalling the ideas if they seem them presented fully.

l'Hôpital's Rule


The reason that L’Hopital’s Rule works is due to the Taylor Series approximation of \begin{align*}f(x)\end{align*} near \begin{align*}x=a\end{align*}. Recall that this formula is given by:

\begin{align*} f(x) = f(a) + f'(a) (x - a) + \frac{f''(a)}{2!}(x - a) ^ 2 + \ldots \end{align*}

and so if \begin{align*}f(a)=g(a)=0\end{align*} we see that near \begin{align*}x=a\end{align*} we have:

\begin{align*}\frac{f(x)}{g (x)} & = \frac{\cancel {f(a)} + f'(a)(x - a) + \frac{f''(a)}{2!} (x - a) ^ 2 + \ldots}{\cancel {g(a)} + g'(a) (x - a) + \frac{g''(a)}{2!} ( x - a) ^ 2 + \ldots}\\ & = \frac{f'(a)(x - a) + \frac{f''(a)}{2!} (x - a) ^ 2 + \ldots}{g'(a) (x - a) + \frac{g''(a)}{2!} ( x - a) ^ 2 + \ldots}\end{align*}

So if we divide top and bottom by the factor \begin{align*}(x - a)\end{align*} we obtain that:

\begin{align*}\frac{f(x)}{g(x)} = \frac{f'(a) + \frac{f'' (a)}{2!}(x - a) + \ldots}{g '(a) + \frac{g'' (a)}{2!} (x - a) + \ldots}\end{align*}

or that, in the limit that \begin{align*}x \rightarrow a\end{align*} :

\begin{align*}\lim_{x \to a}\frac{f(x)}{g(x)} = \frac{f'(a)}{g'(a)} \end{align*}

If this is again an indeterminate the process can be iterated again to obtain second derivatives. Furthermore, if plugging in gave the indeterminate form \begin{align*}\frac{\infty}{\infty}\end{align*}instead of \begin{align*}\frac{0}{0}\end{align*} we could use the same proof by simply examining the limit of the functions \begin{align*}f(q)\end{align*} and \begin{align*}g(q)\end{align*} where \begin{align*}q = \frac{1}{x}\end{align*}.


In an earlier chapter we showed graphically that: \begin{align*}\lim_{h \to 0} \frac{\text {Sin}(h)}{h} = 1\end{align*}

By zooming in near \begin{align*}h=0\end{align*}:

Students now have the means to solve this problems and important ones like it using L’Hopital’s Rule:

\begin{align*}\lim_{h \to 0}\frac{\text {Sin}(h)}{h} = \lim_{h \to 0} \frac{\frac{d}{dh} \text {Sin}(h)}{\frac{d}{dh}h} = \lim_{h \to 0} \frac{\text {Cos}(h)}{1} = 1\end{align*}


To test this material it is sufficient to have students find some limits that are not obvious. However, it is always best to continuously remind them of how the results can often be predicted by making a sketch of the plot near the limiting point. Similarly, it is a good idea to be able to mentally plug in some values near the limit to see if any trend is clearly visible.

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