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# Functions for Calculus

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## Jamon Oxendine-Blackmon

on 21 September 2015

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#### Transcript of Functions for Calculus

Differentiating Functions
By: Jamon Oxendine-Blackmon
Resources
https://www.mathsisfun.com/sets/functions-common.html
http://e.lvme.me/d3ouzpd.jpg
http://www.kappit.com/img/pics/29989393geddg_sm.jpg
http://www.mathsisfun.com/data/function-grapher.php?func1=sqrt(x)&xmin=-12&xmax=12&ymin=-8&ymax=8
Types of Functions
Square Root Function
Absolute Value Function
Sine Function
Reciprical Function
Square Root Function
Absolute Value Fucntion
Sine Function
Reciprocal Function
y=1/(x^3)
y=sqrt(x+4)
y=abs(x-3)
y=sine(2x+7)
y=(3x^3)+(2x^2)+x+1
y=x^-3
d/dx[x^n]=nx^(n-1)
y'=(-3)x^(-3-1)
y'=-3x^-4
y'=-3/(x^-4)
d/dx[x^n]=nx^(n-1)
y'=0.5(x+4)^(0.5-1)
y'=0.5(x+4)^(-0.5)
y'=1/(2sqrt(x+4))
y=(x+4)^(1/2)
abs(x)=sqrt(x^2)
y=sqrt(x-3)
y=(x-3)^(1/2)
d/dx[x^n]=nx^(n-1)

y'=0.5(x-3)^(0.5-1)
y'=1/(2sqrt(x-3))
y'=(1/2)(x-3)^(-0.5)
dy/dx=(dy/du)x(du/dx)
y=sine(u)
u=2x+7
dy/du=cos(u)
du/dx=2
y'=2cos(u)
y'=2cos(2x+7)
d/dx[x^n]=nx^(n-1)
d/dx[c]=0
y'=(3)(3x^(3-1))+(2)(2x^(2-1))+1
d/dx[x]=1
y=(9x^2)+(4x)+1
f(x)=1/(x^3)
f'(x)=3/(x^-4)
f(x)=sqrt(x+4)
f(x)=abs(x-3)
f(x)=sine(2x+7)
f(x)=(3x^3)+(2x^2)+x+1
f'(x)=(9x^2)+(4x)+1
f'(x)=1/(2sqrt(x-3))
f'(x)=2cos(2x+7)
f'(x)=1/(2sqrt(x+4))
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