# The Quotient Rule for Differentiation

The quotient rule provides us with a tool/technique to differentiate functions that can be written as the quotient of two functions, that's one function being divided by another.

We start by stating/learning the formula for the quaotient rule, do make a note of it. We then watch a detailed tutorial illustrating how to use the quotient rule. Finally we work our way through a few exercises to make sure we 've understood how to use this rule. ## Quotient Rule for Differentiation

Given a function $$f(x)$$ that can be written as the quotient of two functions $$u(x)$$ and $$v(x)$$, that's: $f(x) = \frac{u(x)}{v(x)}$ we can differentiate it using the quotient rule: $\text{if} \quad f(x) = \frac{u(x)}{v(x)}$ $\text{then} \quad f'(x) = \frac{u'(x).v(x)-u(x).v'(x)}{\begin{bmatrix} v(x) \end{bmatrix}^2}$ This is further explained and illustrated in tutorial 1 below.

## Exercise 1

1. Find $$\frac{dy}{dx}$$, given $$y=\frac{x^2}{3x+4}$$.
2. Find $$f'(x)$$, given $$f(x) = \frac{x^3}{x^2+8}$$.
3. Differentiate the function defined by $$f(x) = \frac{2x}{e^x}$$.
4. Given $$y=\frac{2x+1}{x^2+x-1}$$, find an expression for $$\frac{dy}{dx}$$.
5. Given $$y=\frac{1+x^2}{1-x^2}$$, find an expression for $$\frac{dy}{dx}$$.
6. Find $$f'(x)$$, given $$f(x)=\frac{3x^2}{tan(x)+1}$$.

1. $$\frac{dy}{dx} = \frac{3x^2+8x}{\begin{bmatrix} 3x + 4 \end{bmatrix}^2}$$
2. $$f'(x) = \frac{x^4 + 24x^2}{\begin{bmatrix} x^2 + 8 \end{bmatrix}^2}$$
3. $$f'(x) = \frac{2 - 2x}{e^x}$$
4. $$\frac{dy}{dx} = -\frac{2x^2 + 2x + 3}{\begin{bmatrix} x^2 + x -1 \end{bmatrix}^2}$$
5. $$\frac{dy}{dx} = \frac{4x}{\begin{bmatrix} 1 - x^2 \end{bmatrix}^2}$$
6. $$f'(x) = \frac{6x.tan(x) + 6x -3x^2.sec^2(x)}{\begin{bmatrix} tan(x) + 1 \end{bmatrix}^2}$$

1. We can write $$y=\frac{x^2}{3x+4}$$ as $$y=\frac{u(x)}{v(x)}$$, where: $u(x)=x^2 \quad \text{and} \quad v(x) = 3x+4$ so: $u'(x)=2x \quad \text{and} \quad v'(x) = 3$ And using the quotient rule we can write: \begin{aligned} \frac{dy}{dx} &= \frac{u'(x).v(x)-u(x).v'(x)}{\begin{bmatrix} v(x) \end{bmatrix}^2} \\ & = \frac{2x(3x+4) - x^2.3}{\begin{bmatrix} 3x + 4 \end{bmatrix}^2} \\ & = \frac{2x.3x+2x.4 - 3x^2}{\begin{bmatrix} 3x + 4 \end{bmatrix}^2} \\ & = \frac{6x^2 + 8x -3x^2}{\begin{bmatrix} 3x + 4 \end{bmatrix}^2} \\ \frac{dy}{dx} & = \frac{3x^2+8x}{\begin{bmatrix} 3x + 4 \end{bmatrix}^2} \end{aligned}

2. We can write $$f(x)= \frac{x^3}{x^2+8}$$ as $$f(x) = \frac{u(x)}{v(x)}$$, where: $u(x) = x^3 \quad \text{and} \quad v(x) = x^2 + 8$ and $u'(x) = 3x^2 \quad \text{and} \quad v'(x) = 2x$ And using the quotient rule we can write: \begin{aligned} f'(x) &= \frac{u'(x).v(x)-u(x).v'(x)}{\begin{bmatrix} v(x) \end{bmatrix}^2} \\ & = \frac{3x^2 \begin{pmatrix} x^2+8 \end{pmatrix} - x^3.2x}{\begin{bmatrix} x^2 + 8 \end{bmatrix}^2} \\ &= \frac{3x^2.x^2 + 3x^2.8 - 2x^4}{\begin{bmatrix} x^2 + 8 \end{bmatrix}^2} \\ & = \frac{3x^4 + 24x^2 - 2x^4}{\begin{bmatrix} x^2 + 8 \end{bmatrix}^2} \\ f'(x) & = \frac{x^4 + 24x^2}{\begin{bmatrix} x^2 + 8 \end{bmatrix}^2} \end{aligned}

3. The function $$f(x) = \frac{2x}{e^x}$$ can be written $$f(x) = \frac{u(x)}{v(x)}$$, where: $u(x) = 2x \quad \text{and} \quad v(x) = e^x$ and $u'(x) = 2 \quad \text{and} \quad v'(x) = e^x$ And using the quotient rule we can write: \begin{aligned} f'(x) &= \frac{u'(x).v(x)-u(x).v'(x)}{\begin{bmatrix} v(x) \end{bmatrix}^2} \\ & = \frac{2e^x - 2xe^x}{\begin{bmatrix} e^x \end{bmatrix}^2 } \\ & = \frac{2e^x - 2xe^x}{ e^{2x} } \\ & = \frac{e^x\begin{pmatrix} 2 - 2x \end{pmatrix}}{ e^x.e^x } \\ f'(x) & = \frac{ 2 - 2x }{ e^x } \end{aligned}

4. We can write $$y= \frac{2x+1}{x^2+x - 1}$$ as $$y = \frac{2x+1}{x^2 + x -1 }$$, where: $u(x) = 2x +1 \quad \text{and} \quad v(x) = x^2 + x -1$ and $u'(x) = 2 \quad \text{and} \quad v'(x) = 2x + 1$ And using the quotient rule we can write: \begin{aligned} \frac{dy}{dx} &= \frac{u'(x).v(x)-u(x).v'(x)}{\begin{bmatrix} v(x) \end{bmatrix}^2} \\ & = \frac{ 2\begin{pmatrix} x^2 + x -1 \end{pmatrix} - \begin{pmatrix} 2x + 1 \end{pmatrix} \begin{pmatrix} 2x + 1 \end{pmatrix} }{ \begin{bmatrix} x^2+x-1 \end{bmatrix}^2} \\ & = \frac{2x^2 + 2x - 2 - \begin{pmatrix} 4x^2 + 4x + 1 \end{pmatrix}}{ \begin{bmatrix} x^2+x-1 \end{bmatrix}^2} \\ & = \frac{2x^2 + 2x -2 - 4x^2 - 4x - 1}{ \begin{bmatrix} x^2 + x - 1 \end{bmatrix}^2} \\ & = \frac{2x^2 + 2x -2 - 4x^2 - 4x - 1}{ \begin{bmatrix} x^2 + x - 1 \end{bmatrix}^2} \\ & = \frac{-2x^2 -2x -3}{\begin{bmatrix} x^2 + x - 1 \end{bmatrix}^2} \\ & = \frac{(-1)\begin{pmatrix}2x^2 +2x +3 \end{pmatrix}}{\begin{bmatrix} x^2 + x - 1 \end{bmatrix}^2}\\ \frac{dy}{dx} & = -\frac{2x^2 +2x +3}{\begin{bmatrix} x^2 + x - 1 \end{bmatrix}^2} \end{aligned}

5. We can write $$y = \frac{1 + x^2}{1 - x^2}$$ as $$y = \frac{u(x)}{v(x)}$$, where: $u(x) = 1 + x^2 \quad \text{and} \quad v(x) = 1 - x^2$ so $u'(x) = 2x \quad \text{and} \quad v'(x) = -2x$ And using the quotient rule we can write: \begin{aligned} \frac{dy}{dx} &= \frac{u'(x).v(x)-u(x).v'(x)}{\begin{bmatrix} v(x) \end{bmatrix}^2} \\ &= \frac{2x \begin{pmatrix} 1 - x^2 \end{pmatrix} - \begin{pmatrix} 1 + x^2 \end{pmatrix} \begin{pmatrix} -2x \end{pmatrix}}{ \begin{bmatrix} 1 - x^2 \end{bmatrix}^2} \\ & = \frac{2x.1 - 2x.x^2 - \begin{pmatrix} 1.\begin{pmatrix}-2x \end{pmatrix} +x^2.\begin{pmatrix} -2x \end{pmatrix} \end{pmatrix}}{ \begin{bmatrix} 1 - x^2 \end{bmatrix}^2 } \\ & = \frac{2x - 2x^3 - \begin{pmatrix} -2x - 2x^3 \end{pmatrix}}{ \begin{bmatrix} 1 - x^2 \end{bmatrix}^2} \\ & = \frac{2x - 2x^3 +2x +2x^3}{ \begin{bmatrix} 1 - x^2 \end{bmatrix}^2} \\ \frac{dy}{dx} & = \frac{4x}{ \begin{bmatrix} 1 - x^2 \end{bmatrix}^2} \end{aligned}

6. We can write $$f(x) = \frac{3x^2}{tan(x)+1}$$ as $$f(x) = \frac{u(x)}{v(x)}$$, where: $u(x) = 3x^2 \quad \text{and} \quad v(x) = tan(x) + 1$ and $u'(x) = 6x \quad \text{and} \quad v'(x) = sec^2(x)$ Using the quotient rule we can write: \begin{aligned} f'(x) &= \frac{u'(x).v(x)-u(x).v'(x)}{\begin{bmatrix} v(x) \end{bmatrix}^2} \\ & = \frac{6x \begin{pmatrix} tan(x) + 1 \end{pmatrix} - 3x^2.sec^2(x) }{ \begin{bmatrix} tan(x) + 1 \end{bmatrix}^2} \\ f'(x) & = \frac{6x . tan(x) + 6x - 3x^2.sec^2(x) }{ \begin{bmatrix} tan(x) + 1 \end{bmatrix}^2} \end{aligned}

## Exercise 2

Given the curve defined by: $y = \frac{ln(x)}{x}$

1. State this function's domain.
2. Find an expression for $$\frac{dy}{dx}$$.
3. Find the equation of the tangent to the curve, when $$x = e$$.

1. $$\text{Domain} = \left \{ x \in \mathbb{R} | x > 0 \right \}$$

2. $$\frac{dy}{dx} = \frac{1-ln(x)}{x^2}$$

3. $$y = \frac{1}{e}$$

We're given the function $$y=\frac{ln(x)}{x}$$.

1. Two conditions need to be met:
• the denominator cannot equal to $$0$$, so $$x \neq 0$$.
• for $$ln(x)$$ to be well-defined we must ensure $$x > 0$$.
So this function's domain is: $\text{Domain} = \left \{ x \in \mathbb{R} | x > 0 \right \}$

2. To find this function's derivative, we use the quotient rule.
Let: $u(x) = ln(x) \quad \text{and} \quad v(x) = x$ Then: $u'(x) = \frac{1}{x} \quad \text{and} \quad v'(x) = 1$ Using the quotient rule: \begin{aligned} \frac{dy}{dx} &= \frac{u'(x).v(x) - u(x).v'(x)}{\begin{bmatrix} v(x) \end{bmatrix}^2} \\ & = \frac{\frac{1}{x}.x - ln(x).1}{x^2} \\ & = \frac{\frac{x}{x} - ln(x)}{x^2} \\ \frac{dy}{dx} & = \frac{1 - ln(x)}{x^2} \end{aligned}

3. We start by finding the $$y$$-coordinate of the curve, when $$x = e$$: $y = \frac{ln(e)}{e} = \frac{1}{e}$ So we're looking for the tangent to the curve at the point $$\begin{pmatrix} e, \frac{1}{e} \end{pmatrix}$$.
We follow the two-step method for finding the tangent:
• Step 1: calculate the gradient of the curve at $$\begin{pmatrix} e, \frac{1}{e} \end{pmatrix}$$.
When $$x = e$$, $$\frac{dy}{dx} = \frac{1 - ln(x)}{x^2}$$ becomes: \begin{aligned} \frac{dy}{dx} & = \frac{1 - ln(e)}{e^2} \\ & = \frac{1 - 1}{e^2} \\ \frac{dy}{dx} &= 0 \end{aligned}

• Step 2: find the equation of the tangent by rearranging the formula $$y-b = m \begin{pmatrix} x -a \end{pmatrix}$$, where $$\begin{pmatrix} a, b \end{pmatrix} = \begin{pmatrix} e, \frac{1}{e} \end{pmatrix}$$ and $$m$$ is the curve's gradient at this point.
In step 1 we found $$\frac{dy}{dx} = 0$$ when $$x = e$$ so $$m=0$$. The formula becomes: \begin{aligned} & y - \frac{1}{e} = 0 \begin{pmatrix} x - e \end{pmatrix} \\ & y - \frac{1}{e} = 0 \\ & y = \frac{1}{e} \end{aligned}
The tangent's equation at the point $$\begin{pmatrix} e, \frac{1}{e} \end{pmatrix}$$ is therefore: $y = \frac{1}{e}$

## Exercise 3

Consider the curve, defined for $$x \neq 0$$, by: $y=\frac{2sin(x)}{x^2}$

1. Find an expression for $$\frac{dy}{dx}$$.
2. Find the value of the gradient of the curve, when $$x = \frac{\pi }{2}$$.
3. Find the equation of the tangent to the curve at the point along its length with $$x$$-coordinate $$x = \frac{\pi}{2}$$.

1. $$\frac{dy}{dx} = \frac{2xcos(x)-4sin(x)}{x^3}$$

2. $$\frac{dy}{dx} = - \frac{32}{\pi^3}$$

3. $$y = - \frac{32}{\pi^3}x + \frac{24}{\pi^2}$$

1. The function $$y = \frac{2sin(x)}{x^2}$$ can be written $$y = \frac{u(x)}{v(x)}$$, where: $u(x) = 2sin(x) \quad \text{and} \quad v(x) = x^2$ and $u'(x) = 2cos(x) \quad \text{and} \quad v'(x) = 2x$ Using the quotient rule, we find $$\frac{dy}{dx}$$: \begin{aligned} \frac{dy}{dx} & = \frac{u'(x).v(x) - u(x).v'(x)}{\begin{bmatrix} v(x) \end{bmatrix}^2} \\ & = \frac{2cos(x).x^2 - 2sin(x).2x}{\begin{bmatrix} x^2 \end{bmatrix}^2} \\ & = \frac{2x^2cos(x) - 4x sin(x)}{x^4} \\ \frac{dy}{dx} & = \frac{2xcos(x) - 4sin(x)}{x^3} \end{aligned}

2. The gradient of the curve when $$x = \frac{\pi }{2}$$ equals the value of the derivative $$\frac{dy}{dx} = \frac{2xcos(x) - 4sin(x)}{x^3}$$ when $$x = \frac{\pi }{2}$$, that's: \begin{aligned} \frac{dy}{dx} & = \frac{2 \times \frac{\pi}{2} \times cos\begin{pmatrix} \frac{\pi}{2} \end{pmatrix} - 4sin\begin{pmatrix} \frac{\pi}{2} \end{pmatrix}}{\begin{pmatrix} \frac{\pi}{2} \end{pmatrix}^3} \\ & = \frac{\pi \times 0 - 4 \times 1}{\frac{\pi^3}{2^3}} \\ & = \frac{-4}{\frac{\pi^3}{8}} \\ & = \frac{-4\times 8}{\pi^3} \\ &= \frac{-32}{\pi^3} \\ \frac{dy}{dx} &= -\frac{32}{\pi^3} \end{aligned}

3. To find the equation of the tangent when $$x = \frac{\pi}{2}$$, we must first calculate the $$y$$-coordinate of the point on the curve when $$x = \frac{\pi }{2}$$.
We do this using the the curve's equation $$y= \frac{2sin(x)}{x^2}$$: \begin{aligned} y & = \frac{2sin\begin{pmatrix} \frac{\pi}{2} \end{pmatrix} }{ \begin{pmatrix} \frac{\pi}{2} \end{pmatrix}^2} \\ & = \frac{2 \times 1 }{ \frac{\pi^2}{2^2}} \\ & = \frac{2 }{ \frac{\pi^2}{4}} \\ & = \frac{2 \times 4 }{\pi^2} \\ y &= \frac{8}{\pi^2} \end{aligned} So the point at which we are looking for the tangent is $$\begin{pmatrix}\frac{\pi}{2}, \frac{8}{\pi^2} \end{pmatrix}$$.

We now use our two-step method for finding the tangent with the formula $$y-b = m \begin{pmatrix} x - a \end{pmatrix}$$, where $$\begin{pmatrix} a, b \end{pmatrix} = \begin{pmatrix}\frac{\pi}{2}, \frac{8}{\pi^2} \end{pmatrix}$$.

• Step 1: find the value of the gradient of the curve at $$x = \frac{\pi}{2}$$.
We found this value in question 2), that was: $\frac{dy}{dx} = -\frac{32}{\pi^3}$ So $$m = -\frac{32}{\pi^3}$$.

• Step 2: find the equation of the tangent by rearranging the formula $$y-b = m \begin{pmatrix} x - a \end{pmatrix}$$: \begin{aligned} & y-b = m \begin{pmatrix} x - a \end{pmatrix} \\ & y - \frac{8}{\pi^2} = -\frac{32}{\pi^3} \begin{pmatrix} x - \frac{\pi}{2} \end{pmatrix} \\ & y - \frac{8}{\pi^2} = -\frac{32}{\pi^3} x + \frac{32}{\pi^3} \times \frac{\pi}{2} \\ & y = -\frac{32}{\pi^3} x + \frac{32}{\pi^3} \times \frac{\pi}{2} + \frac{8}{\pi^2}\\ & y = -\frac{32}{\pi^3}x + \frac{32 \pi}{2 \pi^3} + \frac{8}{\pi^2} \\ & y = -\frac{32}{\pi^3}x + \frac{16 }{\pi^2} + \frac{8}{\pi^2} \\ & y = -\frac{32}{\pi^3}x + \frac{24 }{\pi^2} \end{aligned}
So the tangent's equation at the point $$\begin{pmatrix}\frac{\pi}{2}, \frac{8}{\pi^2} \end{pmatrix}$$ is: $y = -\frac{32}{\pi^3}x + \frac{24 }{\pi^2}$