The Quotient Rule for Differentiation

Answers With Working

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: \[\begin{aligned} f'(x) &= \frac{u'(x).v(x)-u(x).v'(x)}{\begin{bmatrix} v(x) \end{bmatrix}^2} \\ & = \frac{6x \begin{pmatrix} tan(x) + 1 \end{pmatrix} - 3x^2.sec^2(x) }{ \begin{bmatrix} tan(x) + 1 \end{bmatrix}^2} \\ f'(x) & = \frac{6x . tan(x) + 6x - 3x^2.sec^2(x) }{ \begin{bmatrix} tan(x) + 1 \end{bmatrix}^2} \end{aligned}\]

Answers With Working

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: \[y = \frac{1}{e}\]

Answers With Working

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: \[ y = -\frac{32}{\pi^3}x + \frac{24 }{\pi^2} \]