Reflection

This past week we reviewed derivative, absolute extrema, critical numbers, etc. for our 3rd nine week exam. THis is just to review the stuff we have learned.
Example: Find an equation of the tangent line to the graph of f at the given point
f(x) = x/x+4, (-5, 5)
= (x+4)(1) -[(x)(1)] / (x+4)^2
= x+4-x / (x+4)^2
= 4/(x+4)^2
m = 4/(-5+4)^2 = 4/1^2 = 4
point slope: y - 5 = 4(x + 5)

Example for a derivative:
use shortcut
f(x) = x^2
= 2x

Quotient Rule:
f(x)/g(x) = g(x)f’(x) – [f(x)g’(x)] / (g(x))^2
(recopy bottom)(derivative top) – [(recopy top)(derivative bottom)] / (bottom)^2
Example: f(x) = x / x^2 + 1
f’(x) = (x^2 + 1)(1) – [(x)(2x)] / (x^2+1)^2
= x^2 + 1 – 2x^2 / (x^2 + 1)^2
= -x^2 + 1 / (x^2 + 1)^2

STEPS to solving implicit derivatives:
1. Differentiate both sides with respect to x, d_/dx
2. Collect all dy/dx terms on one side, and move the other terms to the other side.
3. Factor out dy/dx.
4. Solve for dy/dx.
5. Simplify, *replace with original equation if possible.**

Word Problems
+ Example:
Air is being pumped into a spherical balloon at a rate of 4.5 cubic feet per
minute. Find the rate of change of the radius when the radius is 2 ft.
(*word problem taken from notes)
1. dv/dt = 4.5 ft^3/min dr/dt = ? r = 2 ft.
2. V = 4/3pir^3
3. dv/dt = 4/3pi[3r^2dr/dt]
= 4pir^2dr/dt
4. 4.5 = 4pi(2)^2dr/dt
4.5 = 16pidr/dt
dr/dt = 9/32pi ft/min.