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Calculus

Higher Derivatives

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.