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We take derivatives of functions. Since the derivative of a function is itself a function, we can take the derivative again. A *higher-order derivative* refers to the repeated process of taking derivatives of derivatives. Higher-order derivatives are applied to sketch curves, motion problems, and other applications.

Notation for higher-order derivatives:

First Derivative | Second Derivative | Third Derivative | Fourth Derivative | Fifth Derivative |

\(\frac{dy}{dx}\) | \(\frac{d}{dx}\left(\frac{dy}{dx}\right)\) | \(\frac{d}{dx}\left(\frac{d}{dx}\left(\frac{dy}{dx}\right)\right)\) | \(\frac{d}{dx}\left(\frac{d}{dx}\left(\frac{d}{dx}\left(\frac{dy}{dx}\right)\right)\right)\) | \(\frac{d}{dx}\left(\frac{d}{dx}\left(\frac{d}{dx}\left(\frac{d}{dx}\left(\frac{dy}{dx}\right)\right)\right)\right)\) |

\(\frac{dy}{dx}\) | \(\frac{d^2y}{dx^2}\) | \(\frac{d^3y}{dx^3}\) | \(\frac{d^4y}{dx^4}\) | \(\frac{d^5y}{dx^5}\) |

\(f'(x)\) | \(f''(x)\) | \(f'''(x)\) | \(f^{(4)}(x)\) | \(f^{(5)}(x)\) |

\(D_xy\) | \(D_x^2y\) | \(D_x^3y\) | \(D_x^4y\) | \(D_x^5y\) |

\(y'\) | \(y''\) | \(y'''\) | \(y^{(4)}\) | \(y^{(5)}\) |

Example 1: Find the third derivative of \[f(x)=\frac{2\pi^2}{6-x}\]

Solution: Instead of performing the quotient rule, we'll simplify the function to \(f(x)=2\pi^2\left(6-x\right)^{-1}\).

\begin{align} f'(x) &=-2\pi^2\left(6-x\right)^{-2}\left(-1\right) = 2\pi^2\left(6-x\right)^{-2} \\ f''(x) &= -4\pi^2\left(6-x\right)^{-3}\left(-1\right) = 4\pi^2\left(6-x\right)^{-3} \\ f'''(x)&= -12\pi^2\left(6-x\right)^{-4}\left(-1\right) = 12\pi^2\left(6-x\right)^{-4} \end{align}

Example 2: Find the second derivative of the equation, \[2xy+y^2=16\]

Solution: Since it will take a few steps to isolate for y, we can use implicit differentiation to find the first derivative with respect to x.

\begin{align} \frac{d}{dx}\left(2xy+y^2\right) &= \frac{d}{dx}16 \\ 2x\frac{dy}{dx}+2y+2y\frac{dy}{dx} &= 0 \\ \frac{dy}{dx}\left(2x+2y\right) &= -2y \\ \frac{dy}{dx} &= \frac{-2y}{2x+2y} = \frac{-y}{x+y} \end{align}

Now we differentiate implicitly again to find the second derivative.

\begin{align} \frac{d^2y}{dx^2} &= \frac{-\frac{dy}{dx}\left(x+y\right)-\frac{dy}{dx}\left(-y\right)}{\left(x+y\right)^2} \\ \frac{d^2y}{dx^2} &= \frac{-\frac{dy}{dx}\left(x+y-y\right)}{\left(x+y\right)^2} \\ \frac{d^2y}{dx^2} &= \frac{-\frac{dy}{dx}\left(x\right)}{\left(x+y\right)^2} \end{align}

We know that \(\frac{dy}{dx} = \frac{-y}{x+y}\) so we can substitute this in and find the second derivative.

\begin{align} \frac{d^2y}{dx^2} &= \frac{-\frac{-y}{x+y}\left(x\right)}{\left(x+y\right)^2} \\ \frac{d^2y}{dx^2} &= \frac{xy}{\left(x+y\right)^3}\end{align}

Example 3: A robotic arm moves according to the displacement \(s\) (in m) equation

\[s=\frac{1}{6}t^4-\frac{7}{6}t^3-2t^2+3\]

where \(t\) is time (in s), for \(t>0\). Determine the times at which the acceleration of the robotic arm will be zero.

Solution: The acceleration is the second derivative of the displacement. Thus, we need to find the second derivative of the equation displayed above. Let \(v\) represent the velocity which is the firster derivative and \(a\) respsent the acceleration.

\begin{align} v=s'&=\frac{2}{3}t^3-\frac{7}{2}t^2-4t \\ a=v'=s''&=2t^2-7t-4 \end{align}

Now we have to solve when \(a=0\)

\begin{align} 0 &=2t^2-7t-4 \\ 0&= (2t+1)(t-4) \\ t&=-\frac{1}{2}, 4\end{align}

We reject \(t=-\frac{1}{2}\) since \(t>0\).

Thus, at 4 seconds, the acceleration of the robotic arm is equal to 0.

Designed by Matthew Cheung. This work is licensed under a Creative Commons Attribution 4.0 International License.

- Last Updated: Apr 19, 2023 1:48 PM
- URL: https://libraryguides.centennialcollege.ca/c.php?g=717032
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