week 30

For this post I will explain some things that people seem to have a lot of trouble with.

For integrating, here are the formulas for trig inverses.

d/dx arcsin(u) = 1/sqrt(1-u^2) * u'

d/dx arccos(u) = -1/sqrt(1-u^2) * u'

d/dx arctan(u) = 1/(1+u^2) * u'

d/dx arccot(u) = -1/(1+u^2) * u'

d/dx arcsec(u) = 1 / (abs(u))(sqrt(u^2-1)) *u'

d/dx arccsc(u) = -1 / (abs(u))(sqrt(u^2-1)) *u'


TANGENT LINE

1. find the x of your original function (usually given).
2. find the y of your original function (plug in x)
3. take derivative of original function.
4. plug x into derived function.
5. set up slope-intercept form.

NORMAL LINE

same thing as tangent line except adding a step after step 4.
the new step 5: make the slope perpendicular. Do this by flipping the fraction and making it negative.

CHAIN RULE

y = (tan(x))^3 + tan(x^3)
y' = 3(tanx)^2*(secx)^2 = sec(x^3)*3x^2

MEAN VALUE THEOREM]

If f is continous on the closed interval [a,b] and differentiable on the open interval (a,b), then there exists a number c in (a,b) such that F'(c) = f(b) - f(a) / b-a

IMPLICIT DERIVATIVE

while taking a derivative, anytime you take y', put a dy/dx in its place. then solve for dy/dx in terms of x.