In the last topic, we created some shortcut rules for finding derivative functions. Let's continue with two very important rules: the Product Rule and the Quotient Rule.
Consider the functions \(f(x)=5x^3\) and \(g(x)=x^2\). Combining into a product function: $$(fg)(x)=5x^3\cdot x^2\implies(fg)(x)=5x^5$$ Using rules from last time, \(f^\prime(x)=15x^2\) and \(g^\prime(x)=2x\), and so $$(f^\prime g^\prime)(x)=15x^2\cdot2x\implies(f^\prime g^\prime)(x)=30x^3$$ However, \(\frac{d}{dx}\left((fg)(x)\right)=25x^4\). This shows that in general, the derivative of the product of two functions DOES NOT equal the product of the two derivatives: $$\frac{d}{dx}\left((fg)(x)\right)\ne\frac{d}{dx}(f(x))\cdot\frac{d}{dx}(g(x))$$ So what does the derivative of a product function look like?
Imagine running on a road at a constant speed. Your speed is determined by two factors: the length of your stride and the number of strides you take each second:
Say your stride length is 3 ft per stride and you take 2 strides per second, then your speed is 6 ft/s. Now suppose your stride length increases by 0.5 ft, from 3 to 3.5 ft. Then the change in speed is calculated as follows:
| change in speed \(=\) change in stride length \(\cdot\) stride rate |
| \(=0.5\cdot2\implies 1\) ft/s |
Alternatively, suppose your stride length remains constant but your stride rate increases by 0.25 stride/s, from 2 to 2.25 stride/s. Then
| change in speed = stride length \(\cdot\) change in stride rate |
| \(=3\cdot0.25\implies 0.75\) ft/s |
If both your stride length and stride rate change simultaneously, we expect two contributions to the change in your running speed:
| change in speed = (change in stride length \(\cdot\) stride rate) \(+\) (stride length \(\cdot\) change in stride rate) |
| \(=1\)ft/s \(+0.75\)ft/s\(\implies1.75\)ft/s |
The argument above correctly illustrates that the derivative (or rate of change) of a product of two functions has two components, defined by the following rule: $$\frac{d}{dx}\left((fg)(x)\right)=f^\prime(x)\cdot g(x)+f(x)\cdot g^\prime(x)$$ I like to remember the Product Rule in words: "the derivative of the first, times the second, plus the first, times the derivative of the second." I am not going to give the traditional limit definition worked out to show you how this Product Rule is formed. A quick Google search will find results showing those details.
Find the derivative of \(h(x)=x^3e^x\).
| Understand that \(f(x)=x^3\) and \(g(x)=e^x\) | \(\implies h^\prime(x)=f^\prime(x)\cdot g(x)+f(x)\cdot g^\prime(x)\) |
| \(\implies h^\prime(x)=(3x^2)(e^x)+(x^3)(e^x)\) | |
| Simplify by factoring out a GCF: | \(\implies h^\prime(x)=x^2e^x\left(3+x\right)\;\color{green}{\checkmark}\) |
It is worth mentioning that for many problems we do, this final step of simplifying is only necessary to check answers in the back of the textbook. For most derivatives, I am going to check if the rules are used correctly.
Now consider a function that is the quotient of two functions: \(h(x)=\large{\frac{f(x)}{g(x)}}\), where \(g(x)\ne0\). What would \(h^\prime(x)\) look like? A clever adaptation and use of the Power Rule generates another useful derivative rule called the Quotient Rule:
\(\large{\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right)=\frac{f^\prime(x)\cdot g(x)-f(x)\cdot g^\prime(x)}{\left(g(x)\right)^2}}\)
Find an equation of the tangent line to \(f(x)=\large{\frac{2x^2}{3x-1}}\) at \(a=1\).
| First find the derivative by the Quotient Rule: | \(\implies f^\prime(x)=\large{\frac{4x(3x-1)+2x^2(3)}{\left(3x-1\right)^2}}\) |
| \(\implies f^\prime(x)=\large{\frac{12x^2-4x+6x^2}{\left(3x-1\right)^2}}\) | |
| \(\implies f^\prime(x)=\large{\frac{18x^2-4x}{\left(3x-1\right)^2}}\) | |
| The point of interest is at \(a=1\), find the \(y\)-coordinate: | \(f(1)=\large{\frac{2(1)^2}{3(1)-1}}\)\(\implies f(1)=\large{\frac{2}{2}}\)\(\implies f(1)=1\) |
| The slope of the tangent line is equal to: | \(f^\prime(1)=\large{\frac{18(1)^2-4(1)}{(3(1)-1)^2}}\)\(\implies f^\prime(1)=\frac{14}{4}\implies f^\prime(1)=\frac{7}{2}\) |
| The equation for the tangent line at \(a=1\) is: | \(y-1=\frac{7}{2}(x-1)\;\color{green}{\checkmark}\) |