\( \newcommand{\abs}[1]{\left| \, {#1} \, \right| } \) \( \newcommand{\cm}{\mathrm{cm} } \) \( \newcommand{\sec}{ \, \mathrm{sec} \, } \) \( \newcommand{\units}[1]{\,\text{#1}} \) \( \newcommand{\vhat}[1]{\,\hat{#1}} \) \( \newcommand{\vhati}{\,\hat{i}} \) \( \newcommand{\vhatj}{\,\hat{j}} \) \( \newcommand{\vhatk}{\,\hat{k}} \) \( \newcommand{\vect}[1]{\boldsymbol{\vec{#1}}} \) \( \newcommand{\norm}[1]{\|{#1}\|} \) \( \newcommand{\arccot}{ \, \mathrm{arccot} \, } \) \( \newcommand{\arcsec}{ \, \mathrm{arcsec} \, } \) \( \newcommand{\arccsc}{ \, \mathrm{arccsc} \, } \) \( \newcommand{\sech}{ \, \mathrm{sech} \, } \) \( \newcommand{\csch}{ \, \mathrm{csch} \, } \) \( \newcommand{\arcsinh}{ \, \mathrm{arcsinh} \, } \) \( \newcommand{\arccosh}{ \, \mathrm{arccosh} \, } \) \( \newcommand{\arctanh}{ \, \mathrm{arctanh} \, } \) \( \newcommand{\arccoth}{ \, \mathrm{arccoth} \, } \) \( \newcommand{\arcsech}{ \, \mathrm{arcsech} \, } \) \( \newcommand{\arccsch}{ \, \mathrm{arccsch} \, } \)

17Calculus - Higher Order Derivatives

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Higher order derivatives are not very difficult. The idea is that, after taking the first derivative, you can take the derivative again to get the second derivative, and so on. Here is a quick example.
Example - - Find the second derivative of \(f(x)=3x^5 + 2x+1\).
The first derivative is \( f'(x) = 15x^4 + 2\).
Taking the derivative again, gives us the second derivative \( f''(x) = 60x^3 \).

Notation - - We can continue taking as many derivatives as we want. The third derivative is written \( f'''(x) \) or \( f^{(3)}(x)\). Notice that the derivative number is written in parentheses so that we can tell we are talking about a derivative and not a power, which we would write \( f^{3}(x)\). Starting with the fourth derivative, we abandon the 'multiple quote' notation and write \(f^{(4)}(x) \) and so on.

Okay, let's work some practice problems calculating higher order derivatives. After that you will be ready to move on to the product rule.

Practice

Unless otherwise instructed, solve these problems giving your answers in completely factored simplified form. Do not use the limit definition to calculate derivatives.

Calculate the first and second derivatives of \(\displaystyle{ f(x) = \frac{x-4}{x} }\)

Problem Statement

Calculate the first and second derivatives of \(\displaystyle{ f(x) = \frac{x-4}{x} }\)

Final Answer

\(\displaystyle{ f'(x) = \frac{4}{x^2} }\), \(\displaystyle{ f''(x) = \frac{-8}{x^3} }\)

Problem Statement

Calculate the first and second derivatives of \(\displaystyle{ f(x) = \frac{x-4}{x} }\)

Solution

3647 video

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Final Answer

\(\displaystyle{ f'(x) = \frac{4}{x^2} }\), \(\displaystyle{ f''(x) = \frac{-8}{x^3} }\)

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Calculate \(f''(x)\) of \(f(x)=3x^4+3x^2+2x \)

Problem Statement

Calculate \(f''(x)\) of \(f(x)=3x^4+3x^2+2x \)

Final Answer

\( f''(x) = 6(6x^2+1) \)

Problem Statement

Calculate \(f''(x)\) of \(f(x)=3x^4+3x^2+2x \)

Solution

The first derivative is \(\displaystyle{\frac{d}{dx}[3x^4+3x^2+2x]=12x^3+6x+2}\)
To find the second derivative, we take the derivative of the first derivative.
\(\displaystyle{\frac{d}{dx}[12x^3+6x+2]=36x^2+6}\)

Final Answer

\( f''(x) = 6(6x^2+1) \)

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For \(\displaystyle{ f(x) = \frac{2x+1}{x-3} }\), calculate
a. \(f(4)\), b. \(f'(4)\), c. \(f''(4)\)
d. slope of the tangent line at \(x=2\)
e. all x-value(s) where \(f'(x)=0\)

Problem Statement

For \(\displaystyle{ f(x) = \frac{2x+1}{x-3} }\), calculate
a. \(f(4)\), b. \(f'(4)\), c. \(f''(4)\)
d. slope of the tangent line at \(x=2\)
e. all x-value(s) where \(f'(x)=0\)

Solution

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You CAN Ace Calculus

Topics You Need To Understand For This Page

Trig Formulas

The Unit Circle

The Unit Circle [wikipedia]

Basic Trig Identities

Set 1 - basic identities

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\)

\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\)

\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\)

\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\)

Set 2 - squared identities

\( \sin^2t + \cos^2t = 1\)

\( 1 + \tan^2t = \sec^2t\)

\( 1 + \cot^2t = \csc^2t\)

Set 3 - double-angle formulas

\( \sin(2t) = 2\sin(t)\cos(t)\)

\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\)

Set 4 - half-angle formulas

\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\)

\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\)

Trig Derivatives

\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\)

 

\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\)

\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\)

 

\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\)

\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\)

 

\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\)

Inverse Trig Derivatives

\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\)

 

\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\)

\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\)

 

\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\)

\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

 

\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

Trig Integrals

\(\int{\sin(x)~dx} = -\cos(x)+C\)

 

\(\int{\cos(x)~dx} = \sin(x)+C\)

\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\)

 

\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)

\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\)

 

\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\)

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Practice Instructions

Unless otherwise instructed, solve these problems giving your answers in completely factored simplified form. Do not use the limit definition to calculate derivatives.

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