Derivative=limf(x+change in x) –f(x) R=x*p P=R-C Change in x Limits:
Point Slope form: y-y,=m(x-x,) Hole (removable discontinuity) Jump: Limit does not exist Vertical Asymptote: Limit does not exist
Walking on graph at x=#, what is the y-value? Find the equation of a tangent line on f(x)=1/x at (1,1) Ex1: lim x^2+4x+3 = (-1)^2 +4(-1) +3 =0 point: (1,1) f(x)=x^-1 m=-1 = -1 = -1 = m x-1 x+1 -1+1 0 m=f’(x) f’(x)=-1x^-2 (1)^2 1
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y-1=-1(x-1) y-1=-x+1 = y=-x+2= EQUATION

Product Rule: f’(x)=u’v+v’u Quotient Rule: f(x)=h(x)g’(x)-g(x)h’(x)=lowd’high-highd’low [h(x)]^2 bottom^2
Chain Rule: derivative of the outside(leave inside alone)*derivative of the inside
Implicit Differentiation: (1) take derivative of each term normally, if term has y on it, we will multiply it by y’
Critical Points: (1) Find f’(x); (2) Set f’(x)=0, solve it; (3) Plot points on # line; (4) Test points around the points in step 3, by plugging them into derivative. If positive: up If negative: down; (5) Write our answer as an interval
Max and Mins (relative extrema): (1) Do all the up and down stuff from 3.1; (2) If you went up then down you have a max; if you went down then up you have a min; (3) Label the points (x,y) for max and mins to get the y, go back to f(x)
Ex2: f(x) =1/4x^4-2x^2 a) Find the open intervals on which the function is increasing or decreasing. f(x) =x^3-4x x=2,-2 0=x(x^2 -4) x(x-2)(x+2)

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How to find inflection points: How to find Intervals of Concavity: (1) Find f’’(x) (1) Find inflection pointsf’’(x)=0 (2) Set Equal to zero and solve (2) Plot points on a # line. Test around them by plugging into f’’(x) (3) Plug into original equation to