Chapter 2.1

Finding the average rate of change in the interval [a,b]

(F(b)-F(a))/(b-a) slope of the secant line

Continuity and differentiability

-F(x)=x^2 x is less than or equal to 0, and x-1 when x is more than 0. Non removable discontinuity at x=0. F'(x)= 2x when x is less than or equal 0, and 1 when x is more than 0. Differentiable? -F(x)=x^2 when x is less than or equal to 0, and x when x is more than 0. F'(x)= 2x when x is less than or equal 0, and 1 when x is more than 0. Continous but not diffrentiable because 2(0) is not equal to 1. -F(x)=3rd root of x. Vertical tangent line is not defined f(x) is not Differentiable at x=0.

d/dx(cosx)=

-Sinx

M=

Change in y/change in x

d/dx (sinx)=

Cosx

Constant multiple rule example

H(x)=5x^2 H'(x)=5(2x)=10x

Tangent line

Hits one point of a curve

Secant line

Hits two points on a curve.

Constant rule

If f(x)=c, then f'(x)=0

Power rule

If f(x)=x^n then f´(x)=nx^(n-1)

Derivative

Is the slope same equation

Find derivative by the limit process

Just plug in everything and simplify to get the equation.

To find an eqation with the derivative

Just plug in point and slope into point-slope formula. Y-Y1=m(X-X1)

For a slope of a tangent line to a curve at point c M=

Lim(under as change in x approaches 0) (f(c+(change in x))-f(c))/change in X.

Power rule applications

Must get the x out of denominator, sometimes you must split the fraction into multiple components. Finding the equation of a tangent line, find the derivative using the power rule, find the slope at the point by plugging the point into the derivative . find the equation using point slope form. Finding points where there are horizontal tangent lines. Slope will be 0 and so will x. Find the derivative, set the derivative equal to 0 and get x=something. Find the point with the original equation and x value you solved for.

Find equation of tangent line when given point and formula

Plug in point and formula and get it down to slope/derivative then do point-slope form

Find the points where the tangent lines are parallel to the secant line

Set the instantaeous = to the average.

Differentiate

Take the derivative

Find k such that the line is tangent to the function

Take the derivative (slope of the curve at point x) and set equal to the slope of the given parallel line. Solve for k. Then set f(x) =tangent line, plug in what you got for k, and solve for x. Go back to the linear function you got for k and sub the x values you just found.

Instananeous rate of change at a point

Use the derivative and plug in the points to find the slope.

Secant line is an approximation of a tangent lines slope.

Want diffence between two points of secant lines to approach 0 to get closer to the tangent line.

How to do it

Will get a point and an equation. Plug (c+change in x) in for x into given equation. C is x in point given and f(c) is y. Calculate equation with that previous substitution then plug that value back into the m equation and solve like regular limit to find slope.

d/dx(cf(x))=

cf'(x)

Sum and difference rules

d/dx(f(x)+/-g(x))=f'(x)+/-g(x)

Other notation

dy/dx or y' d/dx(x^2) take the derivative of x^2