Calc AB Unit 3

chain rule, derivatives of trigonometric/inverse trigonometric functions, derivatives of inverse functions, explicit functions, higher order(multiple) derivatives

chain rule: d/dx [f(g(x))] =

chain rule: d/dx [f(g(x))] =

f’(g(x)) ⋅ g’(x)

d/dx [sin(x)] =

cos(x)

d/dx [cos(x)] =

-sin(x)

d/dx [tan(x)] =

sec²(x)

d/dx [cot(x)] =

-csc²(x)

d/dx [sec(x)] =

sec(x)tan(x)

d/dx [csc(x)] =

-csc(x)cot(x)

d/dx [ln(x)] =

x’/x

d/dx [e^x] =

e^x ⋅ x

d/dx [a^x] =

ln(a) ⋅ a^x

log_a(x) =

1/lna ⋅ ln(x)

when finding dy/dx, what are you solving for?

y’ (derivative of y with respect to x)

derivative of inverse function: (f^-1)’(x) =

1/f’(f^-1(x))

d/dx [arcsin(u)] =

u’/√1-u²

d/dx [arccos(u)] =

-u’/√1-u²

d/dx [arctan(u)] =

u’/1+u²

d/dx [arccot(u)] =

-u’/1+u²

d/dx [arcsec(u)] =

u’/|u|√u²-1

d/dx [arccsc(u)] =

-u’/|u|√u²-1

position function

no derivative

velocity function

first derivative (y’)

acceleration function

second derivative (y’’)