Unit 4, 4.1-4.6

Question 1

Product Rule for derivatives

Answer 1

y’=u’v+uv’
Don’t forget chain rule!

Question 2

Quotient Rule for derivatives

Answer 2

y’ = u’v - uv’/v^2

Question 3

Derivative of
a. sin’x
b. tan’x
c. sec’x
d. cos’x
e. cot’x
f. csc’x

Answer 3

a. cosx
b. sec^2x
c. secxtanx
d. -sinx
e. -csc^2x
f. -cscxcotx

Question 4

Inverse trig functions:
a. y = sin^-1x iff siny=x and y is between ______
b. y = cos^-1 iff cosy=x and y is between ______
c. y = tan^-1 iff tany = x and y is between ______

Answer 4

a. [-pi/2, pi/2]
b. [0, pi]
c. (-pi/2, pi/2)

Question 5

Inverse trig functions:
a. y = cot^-1 iff coty = x and y is between ______
b. y = sec^-1 iff secy=x and y is between ______
c. y = csc^-1 iff cscy=x and y is between ______

Answer 5

a. (0, pi)
b. [0, pi] and x does not equal pi/2
c. [-pi/2, pi/2] and x does not equal 0

Question 6

Inverse function properties
a. If f is invertable, then ______________
b. The graphs f and f^-1 are __________
c. f^-1(x) = __________

Answer 6

a. f(f^-1(x)) = x and f^-1(f(x)) = x
b. mirror images across the line y = x
c. 1/f’(f^-1(x))

Question 7

What is the derivative of (sr stands for square root):
a. sin^-1x and for cos add a ____
b. tan^-1x and for cot add a ____
c. sec^-1x, and for csc add a ____
d. Remember x is the _________ and remember you still have to apply ________

Answer 7

a. 1/sr(1-x^2), add - for cos
b. 1/1+x^2, add - for cot
c. 1/|x|sr(x^2 -1), add - for csc
d. argument, chain rule

Question 8

Fun extra things to know! ((sr is squarerootof)
a. cos^2x + sin^2x =
b. sec^2x-tan^2x =
c. 30 - 60 -90 triangle sides and corresponding pi stuff
d. 45-45-90 triangle sides and corresponding stuff
e. how do you know if a pi something is positive or negative
f. The derivative of a constant is ________ a fun way to check if its a constant is _____

Answer 8

a. 1
b. 1
c. across from 90 is 2x, across from 30 (pi/6) is x and across from 60 (pi/3) is sr 3
d. across from both 45s (pi/4) is 1 and across from 90 is sr 2
e. cast, c is in the 4th quadrant and a is in the 1st and so on
f. 0, if it has a variable (it does not a constant, it doesn’t it is a constant)

Question 9

Steps to solve the example (sr is squarerootof):
Write the equation of the tangent line l(x) to f(x) = sec2x at x =pi/8

Answer 9

1. first find f(pi/8) which gives you sr 2, which means you now know the point is pi/8, sr 2
2. then find f’(x) which is 2sec2xtan2x
3. next find f’(pi/8) which gives you the slope, which is 2 sr2
4. nExt plug the stuff into the equation y- y1 = m(x-x1) which gives you
   y-sr2 = 2sr2(x - pi/8) which simplifies to l(x) = 2sr2(x-pi/8) + sr2

Question 10

if f and g are inverses then point on f (a, b) means that on g there is a point _____

Answer 10

(b, a)

Question 11

a. Function f is differentiable at x = c if _____
b. Function f is differentiable on a interval if _______________
c. Function f is differentiable on a function iff ____________
d. You can tell if something is differentiable by the graph if it is ______

Answer 11

a. f’(c) exists
b. everything in the interval (all x values) is differentiable
c. its differentiable for every value of x in its domain
d. smooth curve, no cusps

Question 12

a. If a function is differentiable at a point it is ______
b. If a function is continuous at a point it does _______
c. If a function is not continuous at a point is does _______

Answer 12

a. cont at that point
b. not mean its differentiable
c. mean its not differentiable

Question 13

Prove: f(x) = x^2 -7x + 7 is cont at x=4
(steps)

Answer 13

1. find f’(x) which is f’(x) = 2x + 7
2. then find f’(4) which is 1
3. f’(4) exists therefore f is cont at x=4

Question 14

Use one-sided limits to find the value(s) of the constants that make the piecewise functions differentiable at x = 2
f(x) = ax^3 when x <_ 2 and b(x-3)^2 + 10 when x>2

Answer 14

1. first put in 2 for x in the og equations which gives you 8a = b + 10
2. next differentiate both equations which gives you 3ax^2 and 2b(x-3)
3. then put 2 into those equations, which gives you 6a = -b or -6a = b
4. then substitute -6a in for b in the first equation which gives you 8a = -6(a) +10 which simplifies to a = 5/7
5. then find b by substituting a in to the second equation, which gives you -6(5/7) = b which simplifies to -30/7

Question 15

a. What does a graph having a vertical tangent line mean
b. What does a graph having a horizontal tangent line mean
c. If the limx->0 f(x) exists that does not mean that _________
d. on a graph thing that was in 4.6 cont + diff is _______ and diff but not cont is always ______

Answer 15

a. that it is not differentiable at the pint where the vertical tangent line is
b. it is differentiable because the slope of that line would be 0
c. f(0) exists
d. the same as just diff, none

Question 16

More fun stuff to remember from last chapter:
a. f(x) = logb x f’(x) = ?
b. f(x) = b^x f’(x) = ?
c. lne or logb b
d. lne^a or logb b^x
e. b^logbx or e^lnx

Answer 16

a. 1/xlnb
b. b^xlnb
c. 1
d. a
e. x

Question 17

solve example f(x) = sec^-1(4x^2)

Answer 17

you should get 2/x sr16x^4 - 1
if you don’t look at notes

Question 18

Given f(2) = 4, f’(2) = 3, f(4) = 6, and
f’(4) = 5 calculate (f^-1)’(4)

Answer 18

1. use equation f^-1(x) = 1/f’(f^-1(x))
2. then plug in 4 cause thats what your finding which gives you f^-1(4) = 1/f’(f^-1(4))
3. we find what f^-1(4) is first, you do that using the a,b b,a thing which means its 2, which gives you1/f’(2)
4. next plug in what f’(2) is, which is 3 which gives you 1/3