# 3.5: Exponentials, Logarithms, and Trig Functions

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## The function $$e^x$$

Recall that exponential functions like $$2^x$$, $$3^x$$, and $$e^x$$. Note that $$e$$ is just a number, equal to about $$2.718$$, and is very special when it is the base of an exponential function. All these exponential functions grow extremely quickly. Here is $$e^x$$, and watch how quickly it flies out of the picture. We can modify it so it doesn’t grow so fast. Consider $$e^{0.1x}$$: But even this starts to grow very quickly when $$x$$ gets large. Here is $$e^{0.1x}$$ again for larger values of $$x$$. Hey, that looks a lot like the graph of $$e^x$$ did! Why is that?

Exponentials at various rates of growth model a wide array of phenomena, including population growth, economic growth, radioactive decay, and more. And the reason $$e^x$$ is a very special function is one of the most amazing formulas in math:

$$\boxed{\cfrac{d}{dx} e^x = e^x}$$

That’s right; $$e^x$$ doesn’t change when you take the derivative!

## The function $$\ln(x)$$

Logarithms, on the other hand, are some of the slowest growing functions. Here is $$\ln(x)$$, which is $$\log_e(x)$$, for large values of $$x$$: Notice even for large values of $$x$$, the function does not get larger than $$4$$ in this picture. Natural log $$\ln(x)$$, which again is the log with base $$e$$, also has a special derivative.

$$\boxed{\cfrac{d}{dx} \ln(x) = \frac{1}{x}}$$

Here is $$\ln(x)$$ in blue plotted with its derivative $$\frac{1}{x}$$ in green. ##### Exponents and Logs and More

Find the following derivatives.

1. $$\frac{d}{dx} (e^x + \ln(x))$$

We just need to take the derivative of each term. The $$e^x$$ stays the same when you take the derivative, so we just leave that piece. The $$\ln(x)$$, as we saw above, has derivative $$\frac{1}{x}$$. Hence

$$\frac{d}{dx} (e^x + \ln(x)) = \boxed{e^x + \frac{1}{x}}.$$

2. $$\frac{d}{dx} (5x^2 + 3e^x)$$

We can use the power rule on $$5x^2$$ — multiply by the two, and subtract one from the two, to get $$10x$$. We then see that $$e^x$$ is $$e^x$$, and the $$3$$ stays along for the ride. So

$$\frac{d}{dx} (5x^2 + 3e^x) = \boxed{10x + 3e^x}.$$

3. $$\frac{d}{dx} \left( \frac{1}{x} + 4\ln(x) \right)$$

Remember that to take the derivative of $$\frac{1}{x}$$, we rewrite as $$x^{-1}$$ and use the power rule, and we have $$-1 x^{-2}$$. For $$4 \ln(x)$$, the $$\ln(x)$$ becomes $$\frac{1}{x}$$, and the four multiplies. Therefore we have

$$\frac{d}{dx} \left( \frac{1}{x} + 4 \ln(x) \right) = -1 x^{-2} + 4\left(\frac{1}{x} \right)$$

But wait — these fraction actually can be added together. First, change the $$x^{-2}$$ back into $$\frac{1}{x^2}$$. Then we’ll get a common denominator, and simplify.

\begin{align*} -1 x^{-2} + 4\left(\frac{1}{x} \right) & = \frac{-1}{x^2} + \frac{4}{x} \\ & = \frac{-1}{x^2} + \frac{4}{x} \cdot \frac{x}{x} \\ & = \frac{-1}{x^2} + \frac{4x}{x^2} \\ & = \frac{4x - 1}{x^2} \end{align*}

Hence we have

$$\frac{d}{dx} \left(\frac{1}{x} + 4 \ln(x)\right) = \boxed{\frac{4x - 1}{x^2}}.$$

Two more quick formulas.

$$\boxed{\cfrac{d}{dx} a^x= \ln(a) \cdot a^x}$$

$$\boxed{\cfrac{d}{dx} \log_a(x) = \frac{1}{\ln(a) \cdot x}}$$

We’ll prove the first rule once we have the chain rule up and running. For the second rule, the proof only requires the base change formula.

##### Log in base $$a$$ proof

Prove the following rule
• $$\cfrac{d}{dx} \log_a(x) = \frac{1}{\ln(a) \cdot x}$$

We use the base change formula, which states

$$\log_a(x) = \frac{\ln(x)}{\ln(a)} = \frac{1}{\ln(a)} \ln(x)$$

From this we see

\begin{align*} \frac{d}{dx} \ \log_a(x) & = \frac{d}{dx} \ \frac{1}{\ln(a)} \ln(x) \\ & = \frac{1}{\ln(a)} \frac{d}{dx} \ \ln(x) \\ & = \frac{1}{\ln(a)} \cdot \frac{1}{x} \\ & = \frac{1}{\ln(a) \cdot x}. \end{align*}

Now this proof demonstrates a tricky thing in calculus: sometimes we can just “bring things out” of the derivative like we did with $$\frac{1}{\ln(a)}$$, and other times we cannot. The reason we could take the $$\frac{1}{\ln(a)}$$ out is that it is considered a constant. The derivative $$\frac{d}{dx}$$ is only measuring the change as $$x$$ changes, not as $$a$$ changes. So $$\frac{1}{\ln(a)}$$ is a constant, or unchanging value as $$x$$ changes. Therefore, by the constant multiple rule, we can just take it out of the derivative.

Now to use the new rules.

##### Base $$a$$ examples

Compute the following derivatives.

1. $$\frac{d}{dx} \ (2^x + 3^x)$$

We use the formula on $$2^x$$ and see the derivative is $$\ln(2) \cdot 2^x$$. Same goes for $$\frac{d}{dx} \ 3^x = \ln(3) \cdot 3^x$$. So the end result is $$\boxed{\ln(2)\cdot 2^x + \ln(3) \cdot 3^x}$$.

2. $$\frac{d}{dx} \ 3\log_2(x)$$

We use the formula to get $$\frac{d}{dx} \log_2(x) = \frac{1}{\ln(2) x}$$, and multiply the answer by $$3$$ and get $$\boxed{\frac{3}{\ln(2) x}}$$.

3. We have

\begin{align*} \frac{d}{dx} \frac{5^x}{\ln(5)} & = \frac{1}{\ln(5)} \frac{d}{dx} 5^x \\ & = \frac{1}{\ln(5)} \ln(5) \cdot 5^x \\ & = \boxed{5^x} \end{align*}

## The functions $$\sin(x)$$ and $$\cos(x)$$

The sine function, denoted $$\sin(x)$$, captures oscillating behavior of waves, circles, pendulums, and more. Cosine, denoted $$\cos(x)$$, is a similar function that does the same thing. Here are $$\sin(x)$$ and $$\cos(x)$$ on a graph. A few important values to know for $$\sin(x)$$ and $$\cos(x)$$ are given by the table below. First note that the $$x$$ values have a $$\pi$$ in them — this is actually a way of measuring angles called radians. You can also use $$\sin$$ and $$\cos$$ with degree angle measurements, but for calculus, it works much better in radians. Make sure your calculator is in radian mode for this class. Notice also that the outputs for $$\sin(x)$$ and $$\cos(x)$$ with $$x = 0$$ are the same as with $$x = 2 \pi$$. That’s not a coincidence — $$\sin$$ and $$\cos$$ functions repeat over and over again every $$\Delta x = 2 \pi$$.

The functions $$\sin(x)$$ and $$\cos(x)$$ work very well with calculus, as shown by these important formulas:

$$\boxed{\cfrac{d}{dx} \ \sin(x) = \ \ \cos(x)}$$
$$\boxed{\cfrac{d}{dx} \ \cos(x) = - \sin(x)}$$

##### Sine and Cosine derivatives

Compute the following:

1. $$\frac{d}{dx} \ -2 \cos(x)$$

Using the formulas from this section and earlier in the chapter, we see

$$\frac{d}{dx} \ -2 \cos(x) = -2 (-\sin(x)) = \boxed{2 \sin(x)}$$

2. $$\frac{d}{dx} (\sin(x) + \cos(x) + e^x + \ln(x))$$

$$\frac{d}{dx} (\sin(x) + \cos(x) + e^x + \ln(x)) = \boxed{\cos(x) - \sin(x) + e^x + \frac{1}{x}}.$$

This page titled 3.5: Exponentials, Logarithms, and Trig Functions is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Tyler Seacrest via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.