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]
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[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]
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[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,