AP Calculus BC Unit 2

derivative at a point

using limits:

- lim [(f(x) - f(c)]/[x-c]

x->c

equation of tangent line:

1. find the derivative at a point

2. plug in points in point slope form (derivative is the slope)

normal line: perpendicular to tangent line

- negative reciprocal slope of tangent line's slope

power rule

f(x) = x^n

f'(x) = nx^n-1

*has to be a VARIABLE raised to a REAL NUMBER (if it is a variable raised to a variable then add ln on both sides to evaluate the derivative after)

product and quotient rule

product rule: h(x) = f(x)g(x)

- h'(x) = f'(x)g(x) + g'(x)f(x)

quotient rule: h(x) = [f(x)/g(x)]

- h'(x) = [f'(x)g(x) - g'(x)f(x)]/[g(x)]^2

*sometimes the function can be combines or simplified to then find the derivative (log rules or trig identities)

chain rule

h(x) = g(f(x))

h'(x) = g'(f(x))f'(x)

*take the derivative going from outside to inside like an onion peel

differentiability

differentiability:

1. determine if the function is continuous

2. determine if lim [f(x) - f(a)]/[x - a] exist

x->a

non-differentiability:

1. function is discontinuous

2. sharp turns/points

3. vertical tangent

*if a function is in absolute value form, must change to piecewise to derive and all equals underneath inequalities go away in the derivative form

trig derivatives

1. f(x) = sin(x), f'(x) = cos(x)

2. f(x) = cos(x), f'(x) = -sin(x)

3. f(x) = tan(x), f'(x) = sec^2(x)

4. f(x) = sec(x), f'(x) = sec(x)tan(x)

5. f(x) = cot(x), f'(x) = -csc^2(x)

6. f(x) = csc(x), f'(x) = -csc(x)cot(x)

inverse trig derivatives

1. f(x) = arcsin(u), f'(x) = [du/dx]/rad(1 - u^2)

2. f(x) = arccos(u), f'(x) = -[du/dx]/rad(1 - u^2)

3. f(x) = arctan(u), f'(x) = (du/dx)/(1 + u^2)

4. f(x) = arccot(u), f'(x) = -(du/dx)/(1 + u^2)

5. f(x) = arcsec(u), f'(x) = [du/dx]/[abs(u) x rad(u^2 - 1)]

6. f(x) = arccsc(u), f'(x) = -[du/dx]/[abs(u) x rad(u^2 - 1)]

implicit differentiation

taking the derivative of an entire equation/function

1. find the derivative

2. if the variable is not the respected variable then add dy/dx

3. solve for dy/dx to find derivative

plugging in:

- (dy/dx) | (x, y) = _______ = derivative value

higher order derivatives

the derivative function sign is the roman numeral for the nth derivative

- find derivative and plug in the pervious derivative value where applicable

- find patterns in higher derivatives to find a big nth derivative faster

*if the highest degree and nth derivative number match the answer will be n!

exponential and logarithmic derivatives

exponential:

- a^u: f'(x) = (a^u)(du/dx)(lna)

- e^u: f'(x) = (e^u)(du/dx)

logarithmic:

- logbase(a)(u): f'(x) = [du/dx]/[ulna]

- ln(u): f'(x) = [du/dx]/u

inverse function derivatives

(f^-1)'(b) = 1/f'(a)

1. make the given b value equal to the non-inverse function to find x values

2. plug x into the derivative function

3. divide 1 by f'(x)

parametric derivatives

finding first derivative:

1. find the derivative of the x and y equations

2. find dy/dx by dividing the two derivatives (the notation of ie. dy/dtheta will cancel out)

finding second derivative:

1. find the derivative of the numerator and denominator of dy/dx separately

2. multiply by the reciprocal of the derivative from the original x equation (same as dividing by the derivative from the x equation)

polar derivatives

1. with the given r equation make the x equation and y equation

- x = rcos(ϑ) & y = rsin(ϑ)

2. find the derivative of each respective equation

3. find dy/dx by dividng the derivative of the y equation by the derivative of the x equation

*when finding a derivative at a theta value notation is always dy/dx | ϑ =__= derivative value

vector derivatives

any parametric equation can be turned into a vector equation (ex. r(t) = < rcosϑ, rsinϑ >

- the first derivative of the vector equation represents the direction of tangent line (sometimes velocity and when dy/dx = 0/0, the particle is at rest)

- second derivative can sometimes be acceleration

*if asked to find speed, it must be the positive magnitude value of velocity