The Double Derivatives of Algebraic Functions

The Double Derivatives of Algebraic Functions

A double derivative or second derivative is simply the derivative of the first derivative. The operation between the first and the second is just the same. The double or second derivative doesn't end the function derivation. It will continue until the function reaches zero. This mathematical function tends to investigate the details of the said function by means of derivation. "Dismantling" might a bit clear when we say we derive the function. To make the discussion more clear, the variable(s) to be derived highlighted in colors.

Illustrative Example:

Find the second derivative of the given functions.

$1.) y=3{\color{Red} x}^{{\color{Blue} 2}}$ , given function

${y}'={\color{Blue} 2}(3{\color{Red} x}^{{\color{Blue} 2-1}})$ , the first derivative, from the rule $\frac{d}{dx}{\color{Red} x}^{{\color{Blue} n}}={\color{Blue} n}{\color{Red} x}^{{\color{Blue} n-1}}$

$y'={\color{Blue} 2}(3{\color{Red} x}^{{\color{Blue} 1}})$ , this is when the power (2-1)

$y'=6{\color{Red} x}$ , this is when the coefficients 2 & 3 multiplied

$y''=6$ , the final answer for the second derivative

$2.) y= 2{\color{Red} x}^{{\color{Blue} 4}}+3{\color{Red} x}^{{\color{Blue} 3}}-{\color{Red} x}^{{\color{Blue} 2}}$ , given function

$y'= 8{\color{Red} x}^{{\color{Blue} 4-1}}+9{\color{Red} x}^{{\color{Blue}3-1}}-2{\color{Red} x}^{{\color{Blue}2-1}}$ , the first derivative, from the rule $\frac{d}{dx}{\color{Red} x}^{{\color{Blue} n}}={\color{Blue} n}{\color{Red} x}^{{\color{Blue} n-1}}$

$y'=8{\color{Red} x}^{{\color{Blue} 3}}+9{\color{Red} x}^{{\color{Blue}2}}-2{\color{Red} x}^{{\color{Blue}1}}$ , when power was performed according to its operator

$y''= 24{\color{Red} x}^{{\color{Blue} 2}} +18{\color{Red} x} -2$ , the final answer for the second derivative

$3.) y= ({\color{Red} x}^{{\color{Blue} 2}}-3)^{{\color{Blue} 2}}$ , given function

$y= ({\color{Red} x}^{{\color{Blue} 4}}-6{\color{Red} x}^{{\color{Blue} 2}}+9)$ , function expansion
$y'=4{\color{Red} x}^{{\color{Blue} 3-1}}-6{\color{Red} x}^{{\color{Blue} 2-1}}+0$ , the first derivative, from the rule    $\frac{d}{dx}{\color{Red} x}^{{\color{Blue} n}}={\color{Blue} n}{\color{Red} x}^{{\color{Blue} n-1}}$

$y'=4{\color{Red} x}^{{\color{Blue} 2}}-6{\color{Red} x}^{{\color{Blue}1}}$

$y''=8{\color{Red} x}^{{\color{Blue} 2}}-6{\color{Red} x}$ , the final answer for the second derivative

Example Problem:

Find the second derivative of the given functions.

$1.) y= 4{\color{Red} x}^{{\color{Blue} 2}}- 2{\color{Red} x}+1$

${y}'= 8{\color{Red} x}-2+0$

${y}{}''=8$ , answer

$2.) y={\color{Red} x}^{{\color{Blue} 3}}-4{\color{Red} x}^{{\color{Blue} 2}}+3{\color{Red} x}$

$y'=3{\color{Red} x}^{{\color{Blue} 2}}-8{\color{Red} x}+3$

$y''=6{\color{Red} x}-8$ , or

$y''=2(3{\color{Red} x}-4)$

$3.) y=\frac{3}{4}{\color{Red} x}^{{\color{Blue} 4}} +\frac{1}{2}{\color{Red} x}^{{\color{Blue} 2}} +3{\color{Red} x}$

$y'=\frac{3}{4}(4{\color{Red} x}^{{\color{Blue} 3}}) +\frac{1}{2}(2{\color{Red} x})+3$

$y'=3{\color{Red} x}^{{\color{Blue} 3}} +{\color{Red} x}+3$

$y''=9{\color{Red} x}^{{\color{Blue} 2}}+1$

$4.)y=\frac{1}{2}(4{\color{Red}x}^{{\color{Blue} 2}}-3)^{{\color{Blue}2}}$ $=\frac{1}{2}(16{\color{Red} x}^{{\color{Blue} 4}}-24{\color{Red} x}^{{\color{Blue} 2}}+9)$

$y=8{\color{Red} x}^{{\color{Blue} 4}}-12{\color{Red} x}^{{\color{Blue} 2}}+\frac{9}{2}$ , reduced

$y'=32{\color{Red} x}^{{\color{Blue} 3}}-24{\color{Red} x}+0$

$y''=96{\color{Red} x}^{{\color{Blue} 2}}-24$

$5.)y=3({\color{Red} x}^{{\color{Blue} 2}}-1)^{{\color{Blue}2}}$ $=3({\color{Red} x}^{{\color{Blue} 4}}-2{\color{Red} x}^{{\color{Blue} 2}}+1)$

$y'=3(4{\color{Red} x}^{{\color{Blue} 3}}-4{\color{Red} x} +0)$

$y'=12{\color{Red} x}^{{\color{Blue} 3}}-12{\color{Red} x}$

$y''=36{\color{Red} x}^{{\color{Blue} 2}}-12$ or

$y''=12(3{\color{Red} x}^{{\color{Blue} 2}}-1)$

Practice Test:

Find the second derivative of the given functions:

$1.)y=3{\color{Red} x}^{{\color{Blue} 4}}-2{\color{Red} x}^{{\color{Blue} 3}}+{\color{Red} x}^{{\color{Blue} 2}}$

$y'=12{\color{Red} x}^{{\color{Blue} 3}}-6{\color{Red} x}^{{\color{Blue} 2}}+2{\color{Red} x}$

$y''=$ click  to answer

$2.) y=\frac{1}{2}({\color{Red} x}+3)^{{\color{Blue} 2}}$    $=\frac{1}{2}({\color{Red} x}^{{\color{Blue} 2}}+6{\color{Red} x} +9)$

$y'=\frac{1}{2}(2{\color{Red} x}+6)$

$y''=$ click to answer

$3.) y=3-{\color{Red} x}^{{\color{Blue} 2}}+2{\color{Red} x}^{{\color{Blue} 3}}-3{\color{Red} x}^{{\color{Blue} 4}}$

$y'=0 -2{\color{Red} x} +6{\color{Red} x}^{{\color{Blue} 2}} -12{\color{Red} x}^{{\color{Blue} 3}}$

$y''=$ click to answer

Simple Quiz:

Derive the given functions up to the second derivative.

$1.)y=2x^{4}-3x^{3}$

$2.)y=\frac{3}{4}(2{\color{Red} x}^{{\color{Blue} 2}}-3)^{{\color{Blue} 2}}$

$3.)y =\frac{1}{2}({\color{Red} t}^{{\color{Blue} 4}}-3{\color{Red} t}^{{\color{Blue} 3}}+{\color{Red} t}-1)$

$4.) y=\frac{1}{({\color{Red} x}+3)^{{\color{Blue} 2}}}$

$5.) y=\frac{3{\color{Red} t}}{(1-{\color{Red} t})}$

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The Double Derivatives of Algebraic Functions

Posted by : Allan_Dell on Saturday, April 18, 2020 | 9:54 PM

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DMG

The Double Derivatives of Algebraic Functions

Posted by : Allan_Dell on Saturday, April 18, 2020 | 9:54 PM

+ comments + 1 comments

Popular posts

Posted by : Allan_Dell on Saturday, April 18, 2020 | 9:54 PM