Quotient Rule – Examples and Practice Problems

Derivation exercises that involve the quotient of functions can be solved using the quotient rule formula. This formula allows us to derive a quotient of functions such as but not limited to $latex \frac{f}{g} (x) = \frac{f(x)}{g(x)}$.

Here, we will look at the summary of the quotient rule. Additionally, we will explore several examples with answers to understand the application of the quotient rule formula.

CALCULUS
Formula for the quotient rule of derivatives 2

Relevant for

Exploring the quotient rule of derivatives with examples.

See examples

CALCULUS
Formula for the quotient rule of derivatives 2

Relevant for

Exploring the quotient rule of derivatives with examples.

See examples

Summary of The Quotient Rule

The quotient rule is a very helpful tool to derive a quotient of functions. It is a rule that states that the derivative of a quotient of two functions is equal to the function in the denominator g(x) multiplied by the derivative of the numerator f(x) subtracted to the numerator f(x) multiplied by the derivative of the denominator g(x), all divided by the square of the denominator g(x).

This gives us the quotient rule formula as:

$$(\frac{f}{g})'(x) = \frac{g(x) \hspace{1.15 pt} \cdot \hspace{1.15 pt} f'(x) \hspace{2.3 pt} – \hspace{2.3 pt} f(x) \hspace{1.15 pt} \cdot \hspace{1.15 pt} g'(x)}{( \hspace{1.15 pt} g(x) \hspace{1.15 pt} )^2}$$

or in a shorter form, it can be illustrated as:

$$\frac{d}{dx}(\frac{u}{v}) = \frac{vu’ \hspace{2.3 pt} – \hspace{2.3 pt} uv’}{v^2}$$

where $latex u = f(x)$ is the numerator/dividend of the given problem and $latex v = g(x)$ is the denominator/divisor of the given problem.

You can use any of these two forms of the product rule formula depending on your preference.

We use this formula to derive functions that have the following form:

$latex \frac{f}{g}(x) = \frac{f(x)}{g(x)}$

or

$latex F(x) = \frac{u}{v}$

where $latex f(x)$ or $latex u$ is the numerator/dividend whilst $latex g(x)$ and $latex v$ is the denominator/divisor of the given problem.


Quotient Rule – Examples with answers

Using the formula detailed above, we can derive various functions that are written as quotients. Each of the following examples has its respective detailed solution. It is recommended for you to try to solve the sample problems yourself before looking at the solution so that you can practice and fully master this topic.

EXAMPLE 1

Find the derivative of $latex f(x) = \frac{x^2}{e^{2x}}$

The first thing we need to do is to list down the quotient rule formula for our reference:

$latex \frac{d}{dx}(\frac{u}{v}) = \frac{vu’-uv’}{v^2}$

Based on our given, we have $latex x^2$ as the numerator/dividend and $latex e^{(2x)}$ as the denominator/divisor.

Based on the quotient rule formula, $latex u$ is the numerator and $latex v$ is the denominator. Therefore, we have

$latex u = x^2$
$latex v = e^{2x}$
$latex f(x) = \frac{u}{v}$

After doing this, we can now use the quotient rule formula to derive our given problem:

$$\frac{d}{dx}(\frac{u}{v}) = \frac{vu’-uv’}{v^2}$$

$$\frac{d}{dx}(\frac{u}{v}) = \frac{v \cdot \frac{d}{dx}(u) – u \cdot \frac{d}{dx}(v)}{v^2}$$

$$\frac{d}{dx}f(x) = \frac{(e^{2x}) \cdot \frac{d}{dx}(x^2) – (x^2) \cdot \frac{d}{dx}(e^{2x})}{(e^{2x})^2}$$

Next, let’s derive $latex u$ and $latex v$ individually and then substitute it to the quotient rule formula later:

$latex u = x^2$
$latex u’ = 2x$
∴ $latex u’$ used the power formula

$latex v = e^{2x}$
$latex v’ = 2e^{2x}$
∴ $latex v’$ used the derivative of exponential functions

By substituting $latex u$, $latex v$, $latex u’$, and $latex v’$ into the quotient rule formula, we have:

$$\frac{d}{dx}f(x) = \frac{(e^{2x}) \cdot (2x) – (x^2) \cdot (2e^{2x})}{(e^{2x})^2}$$

Simplifying algebraically, we get

$$f'(x) = \frac{(2xe^{2x}) – (2x^2 e^{2x})}{(e^{2x})^2}$$

$$f'(x) = \frac{2xe^{2x}}{(e^{2x})^2} – \frac{2x^2 e^{2x}}{(e^{2x})^2}$$

$$f'(x) = \frac{2x}{e^{2x}} – \frac{2x^2}{e^{2x}}$$

And the final answer is:

$$f'(x) = \frac{2x – 2x^2}{e^{2x}}$$

EXAMPLE 2

What is the derivative of $latex f(x) = \frac{ln{(x)}}{\cos{(x)}}$?

It is always advisable to list down the quotient rule formula first for our reference:

$$\frac{d}{dx}(\frac{u}{v}) = \frac{vu’-uv’}{v^2}$$

Based on our given, we have $latex ln{(x)}$ as the numerator/dividend and $latex cos{(x)}$ as the denominator/divisor.

Based on the quotient rule formula, let $latex u$ be the numerator and $latex v$ be the denominator. Therefore, we have

$latex u = \ln{(x)}$
$latex v = \cos{(x)}$
$latex f(x) = \frac{u}{v}$

After doing this, we can now use the quotient rule formula to derive our given problem:

$$\frac{d}{dx}(\frac{u}{v}) = \frac{vu’-uv’}{v^2}$$

$$\frac{d}{dx}(\frac{u}{v}) = \frac{v \cdot \frac{d}{dx}(u) – u \cdot \frac{d}{dx}(v)}{v^2}$$

$$\frac{d}{dx}f(x) = \frac{(\cos{(x)}) \cdot \frac{d}{dx}(\ln{(x)}) – (\ln{(x)}) \cdot \frac{d}{dx}(\cos{(x)})}{(\cos{(x)})^2}$$

Next, let’s derive $latex u$ and $latex v$ individually and then substitute it into the quotient rule formula later:

$latex u = \ln{(x)}$
$latex u’ = \frac{1}{x}$
∴ $latex u’$ used the derivative of logarithmic functions

$latex v = \cos{(x)}$
$latex v’ = -\sin{(x)}$
∴ $latex v’$ used the derivative of trigonometric functions

By substituting $latex u$, $latex v$, $latex u’$, and $latex v’$ into the quotient rule formula, we have:

$$\frac{d}{dx}f(x) = \frac{(cos{(x)}) \cdot (\frac{1}{x}) – (\ln{(x)}) \cdot (-\sin{(x)})}{(\cos{(x)})^2}$$

Simplifying algebraically, we get

$$f'(x) = \frac{(\frac{\cos{(x)}}{x}) – (-\sin{(x)} \ln{(x)})}{(\cos{(x)})^2}$$

$$f'(x) = \frac{\frac{\cos{(x)}}{x}) + \sin{(x)} \ln{(x)}}{(\cos{(x)})^2}$$

$$f'(x) = \frac{\frac{\cos{(x)}}{x}) + \frac{x \sin{(x)} \ln{(x)}}{x}}{\cos^{2}{(x)}}$$

And the final answer is:

$$f'(x) = \frac{\cos{(x)} + x \sin{(x)} \ln{(x)} }{x \cos^{2}{(x)}}$$

EXAMPLE 3

Find the derivative of $latex f(x) = \frac{x^3}{\sin^{2}{(x)}}$

Let’s start by listing down the quotient rule formula for our reference:

$$\frac{d}{dx}(\frac{u}{v}) = \frac{vu’-uv’}{v^2}$$

We have $latex x^3$ as the numerator/dividend and $latex \sin^{2}{(x)}$ as the denominator/divisor.

Letting $latex u$ be the numerator and $latex v$ be the denominator, we have

$latex u = x^3$
$latex v = \sin^{2}{(x)}$
$latex f(x) = \frac{u}{v}$

After doing this, we can now use the quotient rule formula to derive our given problem:

$$\frac{d}{dx}(\frac{u}{v}) = \frac{vu’-uv’}{v^2}$$

$$\frac{d}{dx}(\frac{u}{v}) = \frac{v \cdot \frac{d}{dx}(u) – u \cdot \frac{d}{dx}(v)}{v^2}$$

$$\frac{d}{dx}f(x) = \frac{(\sin^{2}{(x)}) \cdot \frac{d}{dx}(x^3) – (x^3) \cdot \frac{d}{dx}(\sin^{2}{(x)})}{(\sin^{2}{(x)})^2}$$

Next, let’s derive $latex u$ and $latex v$ individually and then substitute it to the quotient rule formula later:

$latex u = x^3$
$latex u’ = 3x^2$
∴ $latex u’$ used the derivative of logarithmic functions

$latex v = \sin^{2}{(x)}$
$latex v’ = 2 \sin{(x)} \cos{(x)}$
∴ $latex v’$ used the derivative of trigonometric functions

By substituting $latex u$, $latex v$, $latex u’$, and $latex v’$ into the quotient rule formula, we have:

$$\frac{d}{dx}f(x) = \frac{(\sin^{2}{(x)}) \cdot (3x^2) – (x^3) \cdot (2 \sin{(x)} \cos{(x)})}{(\sin^{2}{(x)})^2}$$

Simplifying algebraically and applying some trigonometric identities, we get

$$f'(x) = \frac{(3x^2 \sin^{2}{(x)}) – (2x^3 \sin{(x)} \cos{(x)})}{\sin^{4}{(x)}}$$

$$f'(x) = \frac{3x^2 \sin^{2}{(x)}}{\sin^{4}{(x)}} – \frac{2x^3 \sin{(x)} \cos{(x)}}{\sin^{4}{(x)}}$$

$$f'(x) = \frac{3x^2}{\sin^{2}{(x)}} – \frac{2x^3 \cos{(x)}}{\sin^{3}{(x)}}$$

$$f'(x) = 3x^2 \csc^{2}{(x)} – 2x^3 \cdot \frac{\cos{(x)}}{\sin{(x)}} \cdot \frac{1}{\sin^{2}{(x)}}$$

And the final answer is:

$$f'(x) = 3x^2 \csc^{2}{(x)} – 2x^3 \cot{(x)} \csc^{2}{(x)}$$

EXAMPLE 4

What is the derivative of $latex f(x) = \frac{5x^x}{\cos{(3x)}}$?

First, we should list down the quotient rule formula for our reference:

$$\frac{d}{dx}(\frac{u}{v}) = \frac{vu’-uv’}{v^2}$$

Based on our given, we have $latex 5x^x$ as the numerator/dividend and $latex \cos{(3x)}$ as the denominator/divisor.

Therefore, we have

$latex u = 5x^x$
$latex v = \cos{(3x)}$
$latex f(x) = \frac{u}{v}$

After doing this, we can now use the quotient rule formula to derive our given problem:

$$\frac{d}{dx}(\frac{u}{v}) = \frac{vu’-uv’}{v^2}$$

$$\frac{d}{dx}(\frac{u}{v}) = \frac{v \cdot \frac{d}{dx}(u) – u \cdot \frac{d}{dx}(v)}{v^2}$$

$$\frac{d}{dx}f(x) =\frac{(cos{(3x)}) \cdot \frac{d}{dx}(5x^x) – (5x^x) \cdot \frac{d}{dx}(\cos{(3x)})}{(\cos{(3x)})^2}$$

Next, let’s derive $latex u$ and $latex v$ individually and then substitute it into the quotient rule formula later:

$latex u = 5x^x$
$latex u’ = 5x^x \ln{(x)} + 5x^x$
∴ $latex u’$ used implicit differentiation

$latex v = \cos{(3x)}$
$latex v’ = -3 \sin{(3x)}$
∴ $latex v’$ used the derivative of trigonometric functions and the chain rule formula

By substituting $latex u$, $latex v$, $latex u’$, and $latex v’$ into the quotient rule formula, we have:

$$\frac{d}{dx}f(x) = \frac{(\cos{(3x)}) \cdot (5x^x \ln{(x)} + 5x^x) – (5x^x) \cdot (-3 \sin{(3x)})}{(\cos{(3x)})^2}$$

Simplifying algebraically and applying some trigonometric identities, we get

$$f'(x) =\frac{(5x^x \cos{(3x)} \ln{(x)} + 5x^x \cos{(3x)}) – (-15x^x \sin{(3x)})}{\cos^{2}{(3x)}}$$

$$ f'(x) = \frac{5x^x \cos{(3x)} \ln{(x)} + 5x^x \cos{(3x)} + 15x^x \sin{(3x)}}{\cos^{2}{(3x)}}$$

$$f'(x) = \frac{5x^x \cos{(3x)} \ln{(x)}}{\cos^{2}{(3x)}} + \frac{5x^x \cos{(3x)}}{\cos^{2}{(3x)}} + \frac{15x^x \sin{(3x)}}{\cos^{2}{(3x)}}$$

$$f'(x) = 5x^x \ln{(x)} \cdot \frac{1}{\cos{(3x)}} + 5x^x \cdot \frac{1}{\cos{(3x)}} + 15x^x \cdot \frac{\sin{(3x)}}{\cos{(3x)}} \cdot \frac{1}{\cos{(3x)}}$$

And the final answer is:

$$f'(x) = 5x^x \ln{(x)} \sec{(3x)} + 5x^x \sec{(3x)}+ 15x^x \sec{(3x)} \tan{(3x)}$$

EXAMPLE 5

Find the derivative of $latex f'(x) = \frac{x^{e^x}}{e^{\sin{(x)}}}$

As always, we first need to list down the quotient rule formula for our reference:

$$\frac{d}{dx}(\frac{u}{v}) = \frac{vu’-uv’}{v^2}$$

We have $latex x^{e^x}$ as the numerator/dividend and $latex e^{\sin{(x)}}$ as the denominator/divisor. Therefore, we have

$latex u = x^{e^x}$
$latex v = e^{\sin{(x)}}$
$latex f(x) = \frac{u}{v}$

After doing this, we can now use the quotient rule formula to derive our given problem:

$$\frac{d}{dx}(\frac{u}{v}) = \frac{vu’-uv’}{v^2}$$

$$\frac{d}{dx}(\frac{u}{v}) = \frac{v \cdot \frac{d}{dx}(u) – u \cdot \frac{d}{dx}(v)}{v^2}$$

$$\frac{d}{dx}f(x) = \frac{(e^{\sin{(x)}}) \cdot \frac{d}{dx}(x^{e^x}) – (x^{e^x}) \cdot \frac{d}{dx}(e^{\sin{(x)}})}{(e^{\sin{(x)}})^2}$$

Next, let’s derive $latex u$ and $latex v$ individually and then substitute it to the quotient rule formula later:

$latex u = x^{e^x}$
$latex u’ = x^{e^x} e^x \ln{(x)} + \frac{x^{e^x} e^x}{x}$
∴ $latex u’$ used implicit differentiation

$latex v = e^{\sin{(x)}}$
$latex v’ = e^{\sin{(x)}} \cos{(x)}$
∴ $latex v’$ used the derivative of exponential functions

By substituting $latex u$, $latex v$, $latex u’$, and $latex v’$ into the quotient rule formula, we have:

$$\frac{d}{dx}f(x) = \frac{(e^{sin{(x)}}) \cdot (x^{e^x} e^x \ln{(x)} + \frac{x^{e^x} e^x}{x}) – (x^{e^x}) \cdot (e^{\sin{(x)}} \cos{(x)})}{(e^{\sin{(x)}})^2}$$

Simplifying algebraically and applying some trigonometric identities, we get

$$\frac{d}{dx}f(x) = \frac{(e^{\sin{(x)}}) \cdot (x^{e^x} e^x \ln{(x)} + \frac{x^{e^x} e^x}{x})}{(e^{\sin{(x)}})^2}– \frac{(x^{e^x}) \cdot (e^{\sin{(x)}} \cos{(x)})}{(e^{\sin{(x)}})^2}$$

$$\frac{d}{dx}f(x) = \frac{\frac{x^{e^x} (xe^x \ln{(x)} + e^x)}{x} – x^{e^x} \cos{(x)}}{e^{\sin{(x)}}}$$

$$\frac{d}{dx}f(x) = \frac{\frac{x^{e^x} (xe^x \ln{(x)} + e^x) – x^{e^x} (x \cos{(x)})}{x}}{e^{\sin{(x)}}}$$

$$f'(x) = \frac{x^{e^x} (xe^x \ln{(x)} + e^x) – x^{e^x} (x \cos{(x)})}{xe^{\sin{(x)}}}$$

And the final answer is:

$$f'(x) = \frac{x^{e^x} \left[ (e^x (x \ln{(x)} + 1)) – x \cos{(x)} \right]}{xe^{\sin{(x)}}}$$

EXAMPLE 6

What is the derivative of $latex f(x) = \frac{\cos{(x)}}{\sin^{x}{(x)}}$?

For our reference, let’s list down the quotient rule formula first:

$latex \frac{d}{dx}(\frac{u}{v}) = \frac{vu’-uv’}{v^2}$

We have $latex \cos{(x)}$ as the numerator/dividend and $latex \sin^{x}{(x)}$ as the denominator/divisor. Therefore, we have

$latex u = \cos{(x)}$
$latex v = \sin^{x}{(x)}$
$latex f(x) = \frac{u}{v}$

After doing this, we can now use the quotient rule formula to derive our given problem:

$$\frac{d}{dx}(\frac{u}{v}) = \frac{vu’-uv’}{v^2}$$

$$\frac{d}{dx}(\frac{u}{v}) = \frac{v \cdot \frac{d}{dx}(u) – u \cdot \frac{d}{dx}(v)}{v^2}$$

$$\frac{d}{dx}f(x) = \frac{(\sin^{x}{(x)}) \cdot \frac{d}{dx}(\cos{(x)}) – (\cos{(x)}) \cdot \frac{d}{dx}(\sin^{x}{(x)})}{(\sin^{x}{(x)})^2}$$

Next, let’s derive $latex u$ and $latex v$ individually and then substitute it to the quotient rule formula later:

$latex u = \cos{(x)}$
$latex u’ = -\sin{(x)}$
∴ $latex u’$ used the derivative for trigonometric functions

$latex v = \sin^{x}{(x)}$
$latex v’ = x \cot{(x)} \sin^{x}{(x)} + \sin^{x}{(x)} \ln{(\sin{(x)})}$
∴ $latex v’$ used the derivative of trigonometric functions and implicit differentiation

By substituting $latex u$, $latex v$, $latex u’$, and $latex v’$ into the quotient rule formula, we have:

$$\frac{d}{dx}f(x) = \frac{(\sin^{x}{(x)}) \cdot (-\sin{(x)}) – (\cos{(x)}) \cdot (x \cot{(x)} \sin^{x}{(x)} + \sin^{x}{(x)} \ln{(\sin{(x)})})}{(\sin^{x}{(x)})^2}$$

Simplifying algebraically and applying some trigonometric identities, we get

$$\frac{d}{dx}f(x) =\frac{(-\sin{(x)} \sin^{x}{(x)}) – (x \cos{(x)} \cot{(x)} \sin^{x}{(x)} + \cos{(x)} \sin^{x}{(x)} \ln{(\sin{(x)})})}{\sin^{2x}{(x)}}$$

$$f'(x) =\frac{-\sin{(x)} \sin^{x}{(x)} – x \cos{(x)} \cot{(x)} \sin^{x}{(x)} – \cos{(x)} \sin^{x}{(x)} \ln{(\sin{(x)})}}{\sin^{2x}{(x)}}$$

$$f'(x) = \frac{-\sin{(x)} \sin^{x}{(x)}}{\sin^{2x}{(x)}} – \frac{x \cos{(x)} \cot{(x)} \sin^{x}{(x)}}{\sin^{2x}{(x)}} – \frac{\cos{(x)} \sin^{x}{(x)} \ln{(\sin{(x)})}}{\sin^{2x}{(x)}}$$

$$f'(x) = \frac{-\sin{(x)}}{\sin^{x}{(x)}} – \frac{x \cos{(x)} \cot{(x)}}{\sin^{x}{(x)}}– \frac{\cos{(x)} \ln{(\sin{(x)})}}{\sin^{x}{(x)}}$$

$$f'(x) = \frac{-\sin{(x)} – x \cos{(x)} \cot{(x)} – \cos{(x)} \ln{(\sin{(x)})}}{\sin^{x}{(x)}}$$

$$f'(x) = -\frac{\sin{(x)} + x \cos{(x)} \cot{(x)} + \cos{(x)} \ln{(\sin{(x)})}}{\sin^{x}{(x)}}$$

And the final answer is:

$$f'(x) = -\frac{\sin{(x)} + \cos{(x)} \cdot (x \cot{(x)} + \ln{(\sin{(x)})})}{\sin^{x}{(x)}}$$

EXAMPLE 7

Find the derivative of $latex f(x) = \left( \frac{x-3}{\sqrt{x}} \right)^2$

In this specific problem, will use the chain rule and the quotient rule formulas as the two major formulas to derive this problem:

Chain Rule Formula:

$$\frac{d}{dx}(s^n) = ns^{n-1} s’$$

Quotient Rule Formula:

$$\frac{d}{dx}(\frac{u}{v}) = \frac{vu’-uv’}{v^2}$$

The first step we need to do to derive this problem using the chain rule formula since the whole fraction $latex \frac{u}{v}$ is raised to an exponent.

Let

$$s = \frac{u}{v}$$

$$s = \frac{x-3}{\sqrt{x}}$$

$$s^n = \left( \frac{u}{v} \right)^n$$

$$s^n = \left( \frac{x-3}{\sqrt{x}} \right)^2$$

$latex s^n = f(x)$

$$\frac{d}{dx}(s^n) = \frac{d}{dx}(f(x))$$

Then by applying the chain rule formula to our problem, we have

$$ \frac{d}{dx}(s^n) = n \cdot s^{n-1} \cdot \frac{d}{dx}(s)$$

$$\frac{d}{dx}(s^n) = n \cdot \left( \frac{u}{v} \right)^{n-1} \cdot \frac{d}{dx}\left( \frac{u}{v} \right)$$

$$\frac{d}{dx}(s^n) = 2 \cdot \left( \frac{x-3}{\sqrt{x}} \right)^{2-1} \cdot \frac{d}{dx}\left( \frac{x-3}{\sqrt{x}} \right)$$

$$\frac{d}{dx}(f(x)) = 2 \cdot \left( \frac{x-3}{\sqrt{x}} \right) \cdot \frac{d}{dx}\left( \frac{x-3}{\sqrt{x}} \right)$$

The next step we will do is to derive $latex s$, which in this case, we will now use the quotient rule formula. By doing so, we have

Let $latex u$ be the numerator and $latex v$ the denominator of $latex s$. Hence,

$latex u = x-3$
$latex v = \sqrt{x}$
$latex \frac{d}{dx}(s) = \frac{u}{v}$

After doing this, we can now use the quotient rule formula to derive $latex s$:

$$\frac{d}{dx}(s) = \frac{vu’-uv’}{v^2}$$

$$\frac{d}{dx}(s) = \frac{v \cdot \frac{d}{dx}(u) – u \cdot \frac{d}{dx}(v)}{v^2}$$

$$ \frac{d}{dx}(s) = \frac{(\sqrt{x}) \cdot \frac{d}{dx}(x-3) – (x-3) \cdot \frac{d}{dx}(\sqrt{x})}{(\sqrt{x})^2}$$

Next, let’s derive $latex u$ and $latex v$ individually and then substitute it to the quotient rule formula for $latex \frac{d}{dx}(s)$ later:

$latex u = x-3$
$latex u’ = 1$
∴ $latex u’$ used the power rule formulas with sum/difference of derivatives

$latex v = \sqrt{x}$
$latex v = x^{\frac{1}{2}}$
$latex v’ = \frac{x^{-\frac{1}{2}}}{2}$
∴ $latex v’$ used the power rule formula

By substituting $latex u$, $latex v$, $latex u’$, and $latex v’$ into the $latex \frac{d}{dx}(s)$, we have:

$$\frac{d}{dx}(s) = \frac{(x^{\frac{1}{2}}) \cdot (1) – (x-3) \cdot (\frac{x^{-\frac{1}{2}}}{2})}{(x^{\frac{1}{2}})^2}$$

Finally, the last step we need to do is to substitute $latex \frac{d}{dx}(s)$ into $latex \frac{d}{dx}(s^n)$, which is also the $latex \frac{d}{dx}(f(x))$:

$$\frac{d}{dx}(f(x)) = 2 \cdot \left( \frac{x-3}{\sqrt{x}} \right) \cdot \frac{d}{dx}\left( \frac{x-3}{\sqrt{x}} \right)$$

If

$$\frac{d}{dx}(s) = \frac{d}{dx} \left( \frac{x-3}{\sqrt{x}} \right)$$

and

$$\frac{d}{dx}(s) = \frac{(x^{\frac{1}{2}}) \cdot (1) – (x-3) \cdot (\frac{x^{-\frac{1}{2}}}{2})}{(x^{\frac{1}{2}})^2}$$

then we have

$$\frac{d}{dx}(s^n) = 2 \cdot \left( \frac{x-3}{\sqrt{x}} \right) \cdot \left( \frac{(x^{\frac{1}{2}}) \cdot (1) – (x-3) \cdot (\frac{x^{-\frac{1}{2}}}{2})}{(x^{\frac{1}{2}})^2} \right)$$

Simplifying algebraically, we get

$$\frac{d}{dx}(s^n) = 2 \cdot \left( \frac{x-3}{\sqrt{x}} \right) \cdot \left(\frac{x^{\frac{1}{2}} – \frac{x-3}{2x^{\frac{1}{2}}}}{x} \right)$$

$$\frac{d}{dx}(f(x)) = 2 \cdot \left( \frac{x-3}{\sqrt{x}} \right) \cdot \left(\frac{\frac{2x-(x-3)}{2x^{\frac{1}{2}}}}{x} \right)$$

$$f'(x) = 2 \cdot \left( \frac{x-3}{\sqrt{x}} \right) \cdot \left( \frac{x+3}{2x \sqrt{x}} \right)$$

$$f'(x) = \frac{2(x-3)(x+3)}{2x^2}$$

$$f'(x) = \frac{(x-3)(x+3)}{x^2}$$

And the final answer is:

$$f'(x) = \frac{x^2-9}{x^2}$$


Quotient Rule – Practice problems

Solve the following derivation problems and test your knowledge on this topic. Use the quotient rule formula detailed above to solve the exercises. If you have problems with these exercises, you can study the examples solved above.

Find the derivative of $latex f(x) = \frac{x^2+5}{x-4}$

Choose an answer






What is the derivative of $latex f(x) = \frac{\ln{(x)}}{15x^5}$ ?

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Derive $latex y = \frac{4 \tan{(x)}}{1 – \tan^{2}{(x)}}$

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Differentiate $latex \frac{(9x^3+3x)^4}{\sin^{2}{5x}}$

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Find the derivative of $latex \frac{x^x}{\ln{(e^{2^x})}}$

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