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8.11: Quotient Rule and Higher Derivatives

Difficulty Level: At Grade Created by: CK-12
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You may recall hearing about Becky and her Track and Field competition in a prior lesson. Her boyfriend had taken a picture of her just as she started to pull away from the others on the track. We learned how she might learn to identify her instantaneous speed at just the split second the picture was taken by using Calculus to find a derivative.

What if, instead of just finding her speed at that split second, she wanted to find her acceleration?

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- James Sousa: Higher-Order Derivatives


The Quotient Rule

Theorem: (The Quotient Rule) If f and g are differentiable functions at x and g(x) ≠ 0, then
\begin{align*}\frac {d}{dx}\left [ \frac{f(x)}{g(x)} \right ]= \frac {g(x) \frac {d}{dx}\left [{f(x)} \right ] - f(x) \frac{d}{dx} \left [{g(x)} \right ]}{\left [{g(x)} \right ]^2}\end{align*}

In simpler notation

\begin{align*}\left ( \frac{f}{g} \right )'=\frac {g\cdot f'-f\cdot g'}{g^2}\end{align*}

Keep in mind that the order of operations is important (because of the minus sign in the numerator) and \begin{align*}\left ( \frac{f}{g} \right )'\neq \frac {f'}{g'}\end{align*}.

Higher Derivatives

If the derivative \begin{align*}\,\! f'\end{align*} of the function \begin{align*}\,\! f\end{align*} is differentiable, then the derivative of \begin{align*}f'\end{align*}, denoted by \begin{align*}\,\! f''\end{align*} is called the second derivative of \begin{align*}\,\! f\end{align*}. We can continue the process of differentiating derivatives and obtain third, fourth, fifth and higher derivatives of \begin{align*}\,\! f\end{align*}. They are denoted by \begin{align*}\,\! f'\end{align*}, \begin{align*}\,\! f''\end{align*}, \begin{align*}\,\! f'''\end{align*}, \begin{align*}\,\! f^{(4)}\end{align*}, \begin{align*}\,\! f^{(5)}\end{align*}, . . . ,

Example A

Find \begin{align*}= \frac {dy}{dx}\end{align*} for \begin{align*}y = \frac {x^2-5}{x^3+2}\end{align*}


\begin{align*}\frac {dy}{dx}\end{align*} \begin{align*}= \frac {d}{dx}\left [ \frac{x^2-5}{x^3+2} \right ]\end{align*}
\begin{align*}= \frac {(x^3+2).(x^2-5)'-(x^2-5).(x^3+2)'}{(x^3+2)^2}\end{align*}
\begin{align*}= \frac {(x^3+2)(2x)-(x^2-5)(3x^2)}{(x^3+2)^2}\end{align*}
\begin{align*}= \frac {2x^4+4x-3x^4+15x^2}{(x^3+2)^2}\end{align*}
\begin{align*}= \frac {-x^4+15x^2+4x}{(x^3+2)^2}\end{align*}
\begin{align*}= \frac {x(-x^3+15x+4)}{(x^3+2)^2}\end{align*}

Example B

At which point(s) does the graph of \begin{align*}y = \frac {x} {x^2+9}\end{align*} have a horizontal tangent line?


Since the slope of a horizontal line is zero, and since the derivative of a function signifies the slope of the tangent line, then taking the derivative and equating it to zero will enable us to find the points at which the slope of the tangent line is equal to zero, i.e., the locations of the horizontal tangents. Notice that we will need to use the quotient rule here:

\begin{align*}y\end{align*} \begin{align*}= \frac{x} {x^2 + 9}\end{align*}
\begin{align*}y'\end{align*} \begin{align*}= \frac{(x^2 + 9) \cdot f'(x) - x \cdot g' (x^2 + 9)} {(x^2 + 9)^2} = 0\end{align*} \begin{align*}= \frac{(x^2 + 9) (1) - x (2x)} {(x^2 + 9)^2} = 0\end{align*}
Multiply both sides by \begin{align*}(x^2 +9)^2\end{align*},
\begin{align*}x^2 + 9 - 2x^2\end{align*} \begin{align*}= 0\end{align*}
\begin{align*}x^2\end{align*} \begin{align*}= 9\end{align*}
\begin{align*}x\end{align*} \begin{align*}= \pm 3\end{align*}
Therefore, at \begin{align*}x = -3\end{align*} and \begin{align*}x = 3\end{align*}, the tangent line is horizontal.

Example C

Find the fifth derivative of \begin{align*}f(x) = 2x^4 - 3x^3 + 5x^2 - x - 1\end{align*}


To find the fifth derivative, we must first find the first, second, third, and fourth derivatives.

\begin{align*}f '(x)\end{align*} = \begin{align*}8x^3 - 9x^2 + 5x - x\end{align*}
\begin{align*}f ' ' (x)\end{align*} = \begin{align*}24x^2 - 18x + 5\end{align*}
\begin{align*}f ' ' ' (x)\end{align*} = \begin{align*}48x - 18\end{align*}
\begin{align*}f^{(4)} (x)\end{align*} = \begin{align*}48\end{align*}
\begin{align*}f^{(5)} (x)\end{align*} =

Concept question wrap-up

Once Becky has calculated her instantaneous speed at a given point on the track by finding the derivative, she could then take the derivative of that function to find her instantaneous acceleration at the same point in the race.

By finding her instantaneous speed and acceleration at different points in the race, she can learn a lot about what points made a difference in her overall success, and also what points she needs to work on.

Better racing through Calculus!


Instantaneous velocity is speed in a given direction at a single "snapshot" moment in time.

Instantaneous acceleration is change in velocity calculated at a single instant.

Guided Practice


1) Suppose y'(2) = 0 and (y/q)(2) = 0. Find q(2) assuming y(2) = 0

2) Find the derivative of \begin{align*}k(x) = \frac{-2x - 4}{e^x}\end{align*}

3) Given \begin{align*}f(x) = -2x^2 - 4x - 1\end{align*} What is \begin{align*}f''(x)\end{align*}?

4) Given \begin{align*}f(x) = (-x^4 -4x^3 - 5x^2 +3)\end{align*} Find \begin{align*}f''(x)\end{align*} when \begin{align*}x=3\end{align*}


1) Begin with the quotient rule:

\begin{align*}\left(\frac{y}{q}\right)' (2) = \left(\frac{y'(2)q(2) - y(2)q'(2)}{q(2)^2}\right)\end{align*} ..... Substitute
\begin{align*}(0) = \left(\frac{(0)q(2) - (0)q'(2)}{q(2)^2}\right)\end{align*} ..... Substituting again with given values
\begin{align*}0 = \left(\frac{(0)q(2)}{q(2)^2}\right)\end{align*} ..... Simplify with:\begin{align*}(0)q'(2) = 0\end{align*}
\begin{align*}0 = \frac{0}{q(2)}\end{align*}
\begin{align*}q(2) = 0\end{align*}

2) Use the quotient rule: Note: \begin{align*}(-2x - 4)' = -2\end{align*} and \begin{align*}(e^x)' = e^x\end{align*}

\begin{align*}\left(\frac{-2x-4}{e^x}\right)' = \frac{(-2)(e^x) - (-2x - 4)(e^x)}{e^{2x}}\end{align*} ..... Substitute
\begin{align*}\frac{2x + 2}{e^x}\end{align*} ..... Simplify

3) Recall that \begin{align*}f''(x)\end{align*} means "The derivative of the derivative of x"

\begin{align*}f'(x) = f'(-2x^2 - 4x -1) \to f'(x) = -4x -4\end{align*} ..... By the power rule
\begin{align*}f''(x) = f'(-4x -4) = -4\end{align*}
\begin{align*}\therefore f''(x) = -4\end{align*}

4) Recall that \begin{align*}f''(x)\end{align*} means "The derivative of the derivative of x"

\begin{align*}f'(x) = -4x^3 -12x^2 - 10x\end{align*} ..... Use the power rule on f(x)
\begin{align*}f''(x) = -12x^2 - 24x - 10\end{align*} ..... Use the power rule on f'(x)
\begin{align*}f''(3) = -12(3)^2 -24(3) - 10 \to -108 -72 - 10 = -190\end{align*} ..... Substitute 3
\begin{align*}\therefore f''(3) = -190\end{align*}


Use the Quotient Rule to Solve:

  1. Suppose \begin{align*}u'(0) = 98\end{align*} and \begin{align*}(\frac{u}{q})'(0) = 7\end{align*}. Find \begin{align*}q(0)\end{align*} assuming \begin{align*}u(0) = 0\end{align*}
  2. Given: \begin{align*} b(x) = \frac{x^2 - 5x + 4}{-5x + 2}\end{align*} What is: \begin{align*} b'(2)\end{align*}?
  3. Given: \begin{align*}m(x) = \frac {e^x}{3x + 4}\end{align*} what is \begin{align*} \frac{dm}{dx}\end{align*}?
  4. What is \begin{align*}\frac{d}{dx} \cdot \frac{sin (x)}{x - 4}\end{align*}?
  5. \begin{align*}q(x) = \frac{x}{sin (x)}\end{align*}

Solve these Higher Order Derivatives:

  1. Given: \begin{align*}v(x) = -4x^3 + 3x^2 + 2x + 3\end{align*} What is \begin{align*}v''(x)\end{align*}?
  2. Given: \begin{align*}m(x) = x^2 + 5x \end{align*} What is \begin{align*}m''(x)\end{align*}?
  3. Given: \begin{align*}d(x) = 3x^4e^x\end{align*} What is \begin{align*}d''(x)\end{align*}?
  4. Given: \begin{align*}t(x) = -2x^5sin (x)\end{align*} What is \begin{align*}\frac{d^2t}{dx^2}\end{align*}?
  5. What is \begin{align*}\frac{d^2} {dx^2}3x^5e^x\end{align*}?


  1. \begin{align*}y = \frac{3} {\sqrt{x} + 3}\end{align*}
  2. \begin{align*}y = \frac{4x + 1} {x^2 - 9}\end{align*}
  3. Newton’s Law of Universal Gravitation states that the gravitational force between two masses (say, the earth and the moon), m and M is equal two their product divided by the squared of the distance r between them. Mathematically, \begin{align*}F = G\frac{mM} {r^2}\end{align*} where G is the Universal Gravitational Constant (1.602 × 10-11 Nm2/kg2). If the distance r between the two masses is changing, find a formula for the instantaneous rate of change of F with respect to the separation distance r.
  4. Find \begin{align*}\frac{d} {d \psi} \left [\frac{\psi \psi_0 + \psi^3} {3 - \psi_0}\right ]\end{align*} : where \begin{align*}\psi_0\end{align*} is a constant.
  5. Find \begin{align*}\frac{d^3y} {dx^3} |_{x = 1},\end{align*} where \begin{align*}y = \frac{2} {x^3}\end{align*}.

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differentiable A differentiable function is a function that has a derivative that can be calculated.
Instantaneous acceleration The instantaneous acceleration of an object is the change in velocity of the object calculated at a specific point in time.
Instantaneous velocity The instantaneous velocity of an object is the velocity of the object at a specific point in time.
quotient rule In calculus, the quotient rule states that if f and g are differentiable functions at x and g(x) \ne 0, then \frac {d}{dx}\left [ \frac{f(x)}{g(x)} \right ]= \frac {g(x) \frac {d}{dx}\left [{f(x)} \right ] - f(x) \frac{d}{dx} \left [{g(x)} \right ]}{\left [{g(x)} \right ]^2}.

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At Grade
Date Created:
Nov 01, 2012
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