The function [latex]e^x[/latex]
Recall that exponential functions like [latex]2^x[/latex], [latex]3^x[/latex], and [latex]e^x[/latex]. Note that [latex]e[/latex] is just a number, equal to about [latex]2.718[/latex], and is very special when it is the base of an exponential function. All these exponential functions grow extremely quickly. Here is [latex]e^x[/latex], and watch how quickly it flies out of the picture.
We can modify it so it doesn’t grow so fast. Consider [latex]e^{0.1x}[/latex]:
But even this starts to grow very quickly when [latex]x[/latex] gets large. Here is [latex]e^{0.1x}[/latex] again for larger values of [latex]x[/latex].
Hey, that looks a lot like the graph of [latex]e^x[/latex] 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 [latex]e^x[/latex] is a very special function is one of the most amazing formulas in math:
[latex]\boxed{\cfrac{d}{dx} e^x = e^x}[/latex]
That’s right; [latex]e^x[/latex] doesn’t change when you take the derivative!
The function [latex]\ln(x)[/latex]
Logarithms, on the other hand, are some of the slowest growing functions. Here is [latex]\ln(x)[/latex], which is [latex]\log_e(x)[/latex], for large values of [latex]x[/latex]:
Notice even for large values of [latex]x[/latex], the function does not get larger than [latex]4[/latex] in this picture. Natural log [latex]\ln(x)[/latex], which again is the log with base [latex]e[/latex], also has a special derivative.
[latex]\boxed{\cfrac{d}{dx} \ln(x) = \frac{1}{x}}[/latex]
Here is [latex]\ln(x)[/latex] in blue plotted with its derivative [latex]\frac{1}{x}[/latex] in green.
Find the following derivatives.
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[latex]\frac{d}{dx} (e^x + \ln(x))[/latex]
We just need to take the derivative of each term. The [latex]e^x[/latex] stays the same when you take the derivative, so we just leave that piece. The [latex]\ln(x)[/latex], as we saw above, has derivative [latex]\frac{1}{x}[/latex]. Hence
[latex]\frac{d}{dx} (e^x + \ln(x)) = \boxed{e^x + \frac{1}{x}}.[/latex]
-
[latex]\frac{d}{dx} (5x^2 + 3e^x)[/latex]
We can use the power rule on [latex]5x^2[/latex] — multiply by the two, and subtract one from the two, to get [latex]10x[/latex]. We then see that [latex]e^x[/latex] is [latex]e^x[/latex] , and the [latex]3[/latex] stays along for the ride. So
[latex]\frac{d}{dx} (5x^2 + 3e^x) = \boxed{10x + 3e^x}.[/latex]
-
[latex]\frac{d}{dx} \left( \frac{1}{x} + 4\ln(x) \right)[/latex]
Remember that to take the derivative of [latex]\frac{1}{x}[/latex], we rewrite as [latex]x^{-1}[/latex] and use the power rule, and we have [latex]-1 x^{-2}[/latex]. For [latex]4 \ln(x)[/latex], the [latex]\ln(x)[/latex] becomes [latex]\frac{1}{x}[/latex], and the four multiplies. Therefore we have
[latex]\frac{d}{dx} \left( \frac{1}{x} + 4 \ln(x) \right) = -1 x^{-2} + 4\left(\frac{1}{x} \right)[/latex]
But wait — these fraction actually can be added together. First, change the [latex]x^{-2}[/latex] back into [latex]\frac{1}{x^2}[/latex]. 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[latex]\frac{d}{dx} \left(\frac{1}{x} + 4 \ln(x)\right) = \boxed{\frac{4x - 1}{x^2}}.[/latex]
Two more quick formulas.
[latex]\boxed{\cfrac{d}{dx} a^x= \ln(a) \cdot a^x}[/latex]
[latex]\boxed{\cfrac{d}{dx} \log_a(x) = \frac{1}{\ln(a) \cdot x}}[/latex]
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.
- [latex]\cfrac{d}{dx} \log_a(x) = \frac{1}{\ln(a) \cdot x}[/latex]
We use the base change formula, which states
[latex]\log_a(x) = \frac{\ln(x)}{\ln(a)} = \frac{1}{\ln(a)} \ln(x)[/latex]
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 [latex]\frac{1}{\ln(a)}[/latex], and other times we cannot. The reason we could take the [latex]\frac{1}{\ln(a)}[/latex] out is that it is considered a constant. The derivative [latex]\frac{d}{dx}[/latex] is only measuring the change as [latex]x[/latex] changes, not as [latex]a[/latex] changes. So [latex]\frac{1}{\ln(a)}[/latex] is a constant, or unchanging value as [latex]x[/latex] 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.
-
[latex]\frac{d}{dx} \ (2^x + 3^x)[/latex]
We use the formula on [latex]2^x[/latex] and see the derivative is [latex]\ln(2) \cdot 2^x[/latex]. Same goes for [latex]\frac{d}{dx} \ 3^x = \ln(3) \cdot 3^x[/latex]. So the end result is [latex]\boxed{\ln(2)\cdot 2^x + \ln(3) \cdot 3^x}[/latex].
-
[latex]\frac{d}{dx} \ 3\log_2(x)[/latex]
We use the formula to get [latex]\frac{d}{dx} \log_2(x) = \frac{1}{\ln(2) x}[/latex], and multiply the answer by [latex]3[/latex] and get [latex]\boxed{\frac{3}{\ln(2) x}}[/latex].
-
[latex]\frac{d}{dx} \ 5^x / \ln(5)[/latex]
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 [latex]\sin(x)[/latex] and [latex]\cos(x)[/latex]
The sine function, denoted [latex]\sin(x)[/latex], captures oscillating behavior of waves, circles, pendulums, and more. Cosine, denoted [latex]\cos(x)[/latex], is a similar function that does the same thing. Here are [latex]\sin(x)[/latex] and [latex]\cos(x)[/latex] on a graph.
A few important values to know for [latex]\sin(x)[/latex] and [latex]\cos(x)[/latex] are given by the table below.
[latex]\begin{array}{rrr} \mathbf{x} & \mathbf{\sin(x)} & \mathbf{cos(x)} \\ 0 & 0 & 1 \\ \pi/2 & 1 & 0 \\ \pi & 0 & -1 \\ 3 \pi/2 & -1 & 0 \\ 2 \pi & 0 & 1 \end{array}[/latex]
First note that the [latex]x[/latex] values have a [latex]\pi[/latex] in them — this is actually a way of measuring angles called radians. You can also use [latex]\sin[/latex] and [latex]\cos[/latex] 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 [latex]\sin(x)[/latex] and [latex]\cos(x)[/latex] with [latex]x = 0[/latex] are the same as with [latex]x = 2 \pi[/latex]. That’s not a coincidence — [latex]\sin[/latex] and [latex]\cos[/latex] functions repeat over and over again every [latex]\Delta x = 2 \pi[/latex].
The functions [latex]\sin(x)[/latex] and [latex]\cos(x)[/latex] work very well with calculus, as shown by these important formulas:
[latex]\boxed{\cfrac{d}{dx} \ \sin(x) = \ \ \cos(x)}[/latex]
[latex]\boxed{\cfrac{d}{dx} \ \cos(x) = - \sin(x)}[/latex]
Compute the following:
-
[latex]\frac{d}{dx} \ -2 \cos(x)[/latex]
Using the formulas from this section and earlier in the chapter, we see
[latex]\frac{d}{dx} \ -2 \cos(x) = -2 (-\sin(x)) = \boxed{2 \sin(x)}[/latex]
-
[latex]\frac{d}{dx} (\sin(x) + \cos(x) + e^x + \ln(x))[/latex]
[latex]\frac{d}{dx} (\sin(x) + \cos(x) + e^x + \ln(x)) = \boxed{\cos(x) - \sin(x) + e^x + \frac{1}{x}}.[/latex]