Quotient Rule
Deriving Fractions of Functions
In the previous lesson, we have discussed how to derive products of functions, or functions that are multiplied together. How do we go about deriving fractions of functions, or functions that are being divided? We use the Quotient Rule, which is very similar to the product rule, but with an extra complication.
Put into words,
$$\left(\frac{f(x)}{g(x)}\right)’=\frac{(derive\ f)(copypaste\ g)-(copypaste\ f)(derive\ g)}{g\ squared}$$
Notice that the numerator is basically the same as what you’d get for the product rule EXCEPT that we are subtracting instead of adding. We also have that \(g(x)\) squared in the denom. So its not too much extra to memorize.
First, we will let \(f(x)=x^2\) and \(g(x)=x^3+2\). Then, the derivative can be found by substitution as shown below.
Finding Tangent Lines Using the Quotient Rule
Finding tangent lines using the quotient rule works exactly the same as how we do it with any other derivative rule. Nothing changes at all… except the rule we use to find the slope of the tangent line. See example below.
Remember: to find the equation of a tangent line using the point-slope formula for lines \((y-y_0)=m(x-x_0)\), we need three ingredients: \(m\) which is the slope of the line, which can be found from the derivative of your function at the given \(x\)-value, and the \(x\) and \(y\)-value of a point on your line, which we get from the \(x\)-value given, and the \(y\)-value of the function \(H\) at \(x=1\).
To find the slope, we need to find the derivative of \(H\). This was done in the example above. We found that
$$H'(x)=\frac{(2x)(x^3+2)-(x^2)(3x^2)}{[x^3+2]}$$
To see the steps for how it was computed, expand the dropdown box below.
Now, to find \(m\), we need only plug \(x=1\) into \(H'(x)\). This gives us
$$\begin{align}H'(1)&=\frac{(2\cdot(1))((1)^3+2)-((1)^2)(3\cdot(1)^2)}{[(1)^3+2]}\\&=\frac{3}{3}\\&=1\end{align}$$
So we have \(m=1\). Also, we are given the \(x\)-value of the point on the line, namely \(x=1\). We still need a \(y\)-value. This we get by plugging \(x=1\) into \(H\) (NOT \(H’\)). This gives us
$$H(1)=\frac{(1)^2}{(1)^3+2}=\frac{1}{3}$$
So, in summary, we have \(m=1\), \(x_0=1\), and \(y_0=\frac{1}{3}\). Now let’s plug this into the point-slope formula for lines: \((y-y_0)=m(x-x_0)\). This gives us
$$(y-\frac{1}{3})=1\cdot(x-1)$$
When you solve for \(y\), you get the equation of the tangent line
$$y=x-\frac{2}{3}$$.
Done!