AP calc ab unit 3 derivatives

Derivative of sin(x)

Cos(x)

Derivative of cos(x)

-sin(x)

Derivative of tan(x)

Sec²(x)

Derivative of sec(x)

Sec(x)tan(x)

Derivative of csc(x)

-csc(x)cot(x)

Derivative of cot(x)

-csc²(x)

Product rule: (d/dx)(u)(v) =

u’(v) + v’(u)

Quotient rule: (d/dx)/(u/v) =

((u’(v) - (v’)(u))/v²

Chain rule

• derivative of outside function times derivative of inside function

• u’(v) • v’

• f’(g(x)) • g’(x)

Implicit differentiation: d²y/dx²

1. differentiate both sides of the equation with respect to x

2. Collect the terms with dy/dx on one side of the equation

3. Factor out dy/dx

4. Solve for dy/dx

• use quotient rule to find u, u’, v, and v’

• Derivative of y = dy/dx

• Isolate dy/dx

• To find d²y/dx² plug the value for dy/dx back into og derivative

Power rule

ex. d/dx (1/square root: 4x +5)

• d/dx (1/square root: 4x +5) = (4x+5)^(-1/2)

• (-1/2)•(4x+5)^(-3/2) •4

• -2 •(4x+5)^(-3/2)

• -2/1•square root: (4x+5)³

Inverse trig function

(sin^-1) • (square root:2/(2) = theta

Sin(theta) = square root:2/(2)

Inverse trig functions

arcsin(-1/2)=-0.524 theta

• =sin^-1(-1/2)= theta

• Sin(theta)= -1/2

Inverse trig functions

(cos-1)(square root:3/(2)= theta

Cos(theta)= square root:3/(2)

Inverse trig functions

arctan(square root:3) = theta

• = tan^-1(square root:3)=

• Tan(theta)= square root:3

Derivative of arcsine(sin^-1)

Y=(sin^-1) •x

Just the proof for u

• multiply both sides sin

• siny=x/1 - TAKE DERIVATIVE

• cos(y) dy/dx = 1

• Draw a triangle with side lengths as present in what sin(y) equals and use Pythagorean theorem to find cos(y)

• dy/dx = 1/cos(y) = 1/square root: 1-x²

Derivative of arcsine(sin^-1)

When u instead of x

• chain rule

• d/dx(sin^-1)(u) =

• (1/square root:1:u²) •du/dx

Derivative of arctangent(tan^-1)

Y=tan^-1 •x

Just the proof for u

• mutljply both sides by tan

• Tan(y)=x/1

• Take derivative

• sec²y(dy/dx)

• Draw a triangle with side lengths as present in what tan(y) equals and use Pythagorean theorem to find sec²(y)

• dy/dx = 1/sex²y =1/(secy)² = 1/x² +1

Derivative of arctangent (tan^-1)

When u instead of x

• chain rule

• d/dx • (tan^-1)(u) = 1/(1+u²) • du/dx

Derivative of arcsecant (sec^-1)

y=sec^-1 • x

Just the proof for u

• multiply both sides by sec

• secy = x/1 TAKE DERIVATIVE

• Secytany=1

• Draw a triangle with side lengths as present in what sec(y) equals and use Pythagorean theorem to find tan(y)

• dy/dx = 1/secytany = 1/|x|•square root: x²-1

Derivative of arcsecant(sec^-1)

When u instead of x

• chain rule

• d/dx • (sec^-1 •u) = (1/|u| • square root: u²-1) • du/dx

d/dx•e^x=

e^x

d/dx • e^u =

e^u • du/dx

d/dx •a^x =

a^x • ln(a) • du/dx

d/dx •a^u =

a^u • ln • du/dx

d/dx • ln(x) =

ln(x) = log logbase-e^x

d/dx •ln(u)

1/u • du/dx = u’/u

d/dx •log logbase-a^x

1/x•ln(a)

d/dx • log logbase-a (u) =

1/u•ln(a) • du/dx = u’/ u•ln(a)