Lesson Part 1

Recall

We have written quotients of functions as products by applying a negative exponent to the denominator. This allows us to differentiate quotients of functions using the product rule and the chain rule. This method works well for simple functions in the numerator and denominator.

However, is there a rule that will allow us to differentiate \(f(x) = \dfrac{g(x)}{h(x)}\) directly?

In This Module

• We will determine a rule that allows us to differentiate \( f(x) = \dfrac{g(x)}{h(x)} \) directly, without applying negative exponents and the chain rule.
• We will use the technique of converting to products to differentiate a general quotient expression, leading us to the formula for the quotient rule.

The Quotient Rule

Suppose \( f(x) = \dfrac{g(x)}{h(x)}\), where \( g(x)\) and \(h(x) \) are differentiable functions and \( h(x) \neq 0 \).

This equation can be written as \( f(x) = g(x)[h(x)]^{-1} \).

Now we will apply the product rule and chain rule to \(f(x)\) to differentiate this expression.

\[f'(x) = g'(x)[h(x)]^{-1} + g(x)(-1)[h(x)]^{-2}h'(x) \]

Now, let's convert these back to rational expressions by changing the negative exponent back into a positive exponent.

\[f'(x) = \dfrac{g'(x)}{h(x)} - \dfrac{g(x)h'(x)}{[h(x)]^{2}} \]

Simplifying we reach the following expression.

\[f'(x) = \dfrac{g'(x)h(x)-g(x)h'(x)}{[h(x)]^{2}}\]

This is the formula for quotient rule. So if we ever have a function that has a quotient involved, we can use this rule instead of applying the product and chain rule combined.

The Quotient Rule

If \( g(x) \) and \( h(x) \) are differentiable functions with \( h(x) \neq 0 \), with \( f(x) = \dfrac{g(x)}{h(x)} \), then

\[ f'(x) = \dfrac{g'(x)h(x) - g(x)h'(x)}{\left[ h(x) \right]^2} \]

In Leibniz notation, suppose \(y=\dfrac{u}{v}\) where \(u\) and \(v\) are differentiable functions and \( v \neq 0 \), then

\[\dfrac{dy}{dx}=\dfrac{v~\dfrac{du}{dx}-u~\dfrac{dv}{dx}}{v^2}\]

Note:
When using the quotient rule, the order of the terms in the numerator is very important. We must make sure that we follow this formula accurately.

Lesson Part 2

Let's now try to apply the quotient rule to differentiate the following function.

Examples

Example 1

Differentiate \( f(x) = \dfrac{3x - 4}{x^2 + 5} \). For this function, \( g(x) = 3x - 4 \) and \(h(x) = x^2 + 5 \).

Solution

To apply the quotient rule, we must differentiate these equations, doing so gives \( g'(x) = 3 \) and \( h'(x) = 2x \). Now we can substitute them into the formula.

The formula for the quotient rule states

\[f'(x) = \dfrac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2}\]

We substitute in the functions we found earlier.

\[f'(x)= \dfrac{(3)(x^2 + 5) - (3x - 4)(2x)}{(x^2 + 5)^2} \]

Then we expand the numerator while keeping the denomiantor factored.

\[f'(x) = \dfrac{3x^2 + 15 - 6x^2 + 8x}{(x^2 + 5)^2}\]

Lastly, we simplify the numerator to find the equation of the derivative.

\[f'(x) = \dfrac{-3x^2 + 8x + 15}{(x^2 + 5)^2}\]

It is best to keep the denominator factored in this case. In upcoming modules where we're going to be applying the quotient rule to help us do things like curve sketching or optimization, it is usually more useful to have the denominator in factored form rather than in fully expanded form. So let's just keep it as it is.

Example 2

Let's now try the next example by applying the quotient rule to differentiate \(y\) with respect to \(x\).

Find the point(s) on the graph of \(y=\dfrac{3x}{x+4}\) where the slope of the tangent to the curve is \(\dfrac{12}{25}.\)

Give it a try now.

Solution

Since \(y\) has a quotient where the numerator is the function \(3x\) and the denominator is the function \(x + 4\), let's apply the quotient rule to differentiate this expression.

\[ \begin{align*} \dfrac{dy}{dx} &= \dfrac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2} \\ &= \dfrac{(3)(x+4) - (3x)(1)}{(x+4)^2} \end{align*} \]

Let's multiply the two terms in the numerator separately, and then subtract.

\[\dfrac{dy}{dx} = \dfrac{3x+12 - 3x}{(x+4)^2} \]

Simplifying, we get the following equation.

\[ \dfrac{dy}{dx} = \dfrac{12}{(x+4)^2} \]

Since the desired slope of the tangent to the curve is \(\dfrac{12}{25} \), we know that \(\dfrac{dy}{dx}=\dfrac{12}{25}\). Let's substitute that into our equation for the derivative.

\[\dfrac{12}{25} = \dfrac{12}{\left(x+4\right)^2} \]

Once we have substituted, we can either use cross-multiplication or the idea of the inverse to solve for \(x\).

\[ \begin{align*} 12(x+4)^2 &= 12(25) \\ (x+4)^2 &= 25 \\ x &= \pm\sqrt{25}-4 \\ x &= \pm5-4 \end{align*} \]

So, \(x=1\) or \(x=-9\) at the points on \(y=\dfrac{3x}{x+4}\) where \(\dfrac{dy}{dx}=\dfrac{12}{25}\).

Let's now solve for their corresponding \(y\) values to get the \(x\) and \(y\)-coordinate for both points.

Solving for the corresponding \(y\)-values, we have the two points \(\left(1,\frac{3}{5}\right)\) and \(\left(-9,\frac{27}{5}\right)\).

Example 3

Let's try one more example.

Determine \(\dfrac{dy}{dx}\) if \( y = \dfrac{8\sqrt{x}-1}{x^3 + 1} \) using the quotient rule.

Solution

Before differentiating this expression, rather than having \(\sqrt{x}\), let's apply \(\dfrac{1}{2}\) as an exponent to \(x\), and then differentiate using our power rule.

\[y = \dfrac{8x^{\frac{1}{2}}-1}{x^3 + 1}\]\[\dfrac{dy}{dx} = \dfrac{(4x^{-\frac{1}{2}})(x^3 + 1) - (8\sqrt{x}-1)(3x^2)}{(x^3 + 1)^2} \]

Again, let's multiply the first and second terms in the numerator separately, and then perform the subtraction. In this case we're going to have multiple rational exponents.

\[\dfrac{dy}{dx} = \dfrac {4x^{\frac{5}{2}} + 4x^{-\frac{1}{2}} - 24x^2\sqrt{x}+3x^2}{(x^3 + 1)^2} \]

Let's now simplify some of these rational exponents by the method of rationalizing. Let's multiply this expression by \(\dfrac{\sqrt{x}}{\sqrt{x}}\).

\[\dfrac{dy}{dx} =\dfrac{\sqrt{x}}{\sqrt{x}}\left[ \dfrac {4x^{\frac{5}{2}} + 4x^{-\frac{1}{2}} - 24x^2\sqrt{x}+3x^2}{(x^3 + 1)^2}\right] \]

Simplifying these terms, we have the answer

\[\dfrac{dy}{dx} =\dfrac {-20x^{3} +3x^2\sqrt{x} + 4}{\sqrt{x}(x^3 + 1)^2}\]

Challenge Question

Try this on your own and then see the solution.
Show that there are no tangents to the curve \(f(x)=\dfrac{5x+2}{x+2}\) that have a negative slope.

Solution

We will show that \(f'(x) \gt 0\) for all \(x\) in the domain of \(f(x)\).

Note that the domain of \(f(x)\) is \(\{x\in\mathbb{R} \mid x \neq -2\}\).

\[ \begin{align*} f'(x) &= \dfrac{5(x+2)-(5x+2)(1)}{(x+2)^2} \\ &= \dfrac{5x+10-5x-2}{(x+2)^2} \\ &= \dfrac{8}{(x+2)^2} \end{align*} \]

So, \(f'(x)=\dfrac{8}{(x+2)^2}\).

Since \(8 \gt 0\) and \((x+2)^2 \gt 0\) for all values of \(x \neq -2\), then \(f'(x)\gt 0\) for all \(x\) in the domain of \(f(x)\).

Therefore, every tangent to the curve \(f(x)=\dfrac{5x+2}{x+2}\) has positive slope.

Quiz

See the quiz in the side navigation.