3.5: Exponentials, Logarithms, and Trig Functions
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.
Find the following derivatives.
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\(\frac{d}{dx} (e^x + \ln(x))\)
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
\(\frac{d}{dx} (e^x + \ln(x)) = \boxed{e^x + \frac{1}{x}}.\)
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\(\frac{d}{dx} (5x^2 + 3e^x)\)
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
\(\frac{d}{dx} (5x^2 + 3e^x) = \boxed{10x + 3e^x}.\)
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\(\frac{d}{dx} \left( \frac{1}{x} + 4\ln(x) \right) \)
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.
- \(\cfrac{d}{dx} \log_a(x) = \frac{1}{\ln(a) \cdot x}\)
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.
Compute the following derivatives.
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\(\frac{d}{dx} \ (2^x + 3^x)\)
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 \(\boxed{\ln(2)\cdot 2^x + \ln(3) \cdot 3^x}\).
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\(\frac{d}{dx} \ 3\log_2(x)\)
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 \(\boxed{\frac{3}{\ln(2) x}}\).
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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:
\(\boxed{\cfrac{d}{dx} \ \sin(x) = \ \ \cos(x)}\)
\(\boxed{\cfrac{d}{dx} \ \cos(x) = - \sin(x)}\)
Compute the following:
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\(\frac{d}{dx} \ -2 \cos(x)\)
Using the formulas from this section and earlier in the chapter, we see
\(\frac{d}{dx} \ -2 \cos(x) = -2 (-\sin(x)) = \boxed{2 \sin(x)}\)
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\(\frac{d}{dx} (\sin(x) + \cos(x) + e^x + \ln(x))\)
\(\frac{d}{dx} (\sin(x) + \cos(x) + e^x + \ln(x)) = \boxed{\cos(x) - \sin(x) + e^x + \frac{1}{x}}.\)