Question

Chain Rule

Answer 1

Is this the chain rule one

Answer 2

yes but pay attention to what the question is actually asking

a, b, and c do not require the chain rule as they are not compositions of functions

Answer 3

a, b, and c are just straightforward derivatives of f, g, and h separately

for d) you will apply the chain rule to f(h(x))
and for e) you will apply the chain rule to f(g(x))

Answer 4

derivative of f(x) woild be 2x?

Answer 5

because 1 is constant so 0

Answer 6

no, the square root makes a big difference here


f(x) = (x^2+1)^(1/2)
f'(x) = (1/2)(x^2+1)^(-1/2)*(2x)

Answer 7

wait where's the hLF?

Answer 8

isn't it just sqrt x^2+1

Answer 9

sqrt (x^2+1) = (x^2+1)^(1/2)

Answer 10

my reasoning for re-writing

sqrt (x^2+1)

as

(x^2+1)^(1/2)

makes it easier to take the derivative using the power rule

Answer 11

GOt it

Answer 12

the derivative for sin 3x would be 3xsin?

Answer 13

(3x) sin ^2

Answer 14

have to apply chain rule

for sin(A), change sin to cos, then multiply by the derivative of A

so derivative of sin(x^3) --> cos(x^3) * the derivative of x^3

Answer 15

oh okay so the dr of sin x^3 is cos x^3

Answer 16

"cos(x^3) * the derivative of x^3"

so not just cos(x^3), cos(x^3) * 3x^2

Answer 17

oh oksy

Answer 18

depending on what you've covered in class you may want to add at least the first three to your cheat sheet

Answer 19

okay and when i find the derivaitver its like the = whatever * derivative of x as the answer right

Answer 20

* derivative of what's inside the trig function, yes

Answer 21

so like, d/dx sin(A) = cos(A) * A'

Answer 22

it won't always be x, gotta apply to what you are given

Answer 23

right

Answer 24

can we just do maybe one or two more examples of the chain rule, not 100% on it yet :S

Answer 25

have you tried the other problems on your WS?

Answer 26

OH forgot about those okay so for d :

Answer 27

f(h(x)) = it would jutbe (1/2)(x^2+1)^-1/2 (2x)

Answer 28

f(h(x)) derivative ---> f'(h(x)) * h'(x)

Answer 29

it would just be one inside the other right

Answer 30

f'(x) = (1/2)(x^2+1)^(-1/2)*(2x) as we calculated from step a)

f'(h(x)) is therefore (1/2)([e^(-x^2/2)]^2+1)^(-1/2)*(2[e^(-x^2/2)]^2)

then you will have to multiply this whole expression by the derivative of h(x)

Answer 31

UGh not understanding how you got the equation f'(h(x))

Answer 32

it's part of the definition of the chain rule

Answer 33

Still here just writing notes

Answer 34

\(\color{#0cbb34}{\text{Originally Posted by}}\) @Vocaloid
f'(x) = (1/2)(x^2+1)^(-1/2)*(2x) as we calculated from step a)

f'(h(x)) is therefore (1/2)([e^(-x^2/2)]^2+1)^(-1/2)*(2[e^(-x^2/2)]^2)

then you will have to multiply this whole expression by the derivative of h(x)
\(\color{#0cbb34}{\text{End of Quote}}\)

Answer 35

SO this is the definition of chain rule

Answer 36

\(\color{#0cbb34}{\text{Originally Posted by}}\) @Vocaloid
f(h(x)) derivative ---> f'(h(x)) * h'(x)
\(\color{#0cbb34}{\text{End of Quote}}\)

this is the chain rule

Answer 37

the next post is just me going through the steps to calculate exactly what f'(h(x)) * h'(x) is

Answer 38

Okay can we do e

Answer 39

It would be

Answer 40

f'(g(x)) * g'(x)

f'(g(x)) * g'(x) is (1/2)(x^2+1)^-1/2(2x) * (sin(x^3) * (cos x ^3*3x^2)

Answer 41

almost

"f'(g(x)) "

so you have to plug g(x) into x for the derivative of f(x)

Answer 42

f'(x) = (1/2)(x^2+1)^-1/2(2x)

so f'(g(x)) = (1/2)(g(x)^2+1)^(-1/2)(2*g(x)))

then replace "g(x)" with the sin(x^3) since that's what g(x) is

Answer 43

then all of that * (cos x ^3*3x^2)

Answer 44

so f'g(x)) would be f prime along with g of x

Answer 45

f'(g(x)) means take the derivative of f(x), then replace all the x's with g(x)

Answer 46

okay next is implicit differentiation

Answer 47

Can we try another example I am not getting the pictutre :S

Answer 48

you just take the derivative of both sides w.r.t x as normal

if you have a mixed term like xy, you just apply the product rule to get

x'y + y'x

since you're solving for dy/dx (in other words, y') simplify everything except y' and then at the end, solve for y'

Answer 49

intermeidate value theornm

Answer 50

Ughim falling aaslpee

Answer 51

if a function is continuous between a and b such that

f(a) < N < f(b)

then there must be some value c in between a and b such that f(c) = N

Answer 52

another way to think about it: if the function is continuous then it must hit every possible value between f(a) and f(b)

Answer 53

example ?

Answer 54

let's say f(x) is continuous between [1,2]
f(1) = 1
f(2) = 2

we know there must be some x-value such that f(x) = 1.5 since 1.5 is between the two function values

Answer 55

oh thats all its just picking a value in between? and showing it

Answer 56

yes, the IVT lets us prove that a certain x-value exists for a desired y-value

Answer 57

could htis be possible with 1.3

Answer 58

or maybe even 1.7

Answer 59

and the value we come up with is y?

Answer 60

what would be the x

Answer 61

yes, 1.3 and 1.7 are both within the bounds

the IVT doesn't let us solve for x, it just lets us know that there's an x-value that exists for that y-value

Answer 62

okay next is mean value theorem

Answer 63

can open a new post? if you want

Answer 64

tbh i don't know how to use this practically, i just know that this exists

Answer 65

Thats okay I'll study that one myself

Answer 66

Linear and Quadratic Approx

Answer 67

I have a handout for these

Answer 68

yeah you've pretty much got it

Answer 69

Copied these from what my professor wrote :x

Answer 70

Idk what I'm doing

Answer 71

just stick the formulas for linear and quadratic formulas on your cheat sheet

Answer 72

:(

Answer 73

linear approximation

y = f(a) + f'(a) * (x-a) at f(x) for x = a

applying this logic to f(x) = ln(x) at x = 1

y = ln(1) + ln'(1) * (x-1)

Answer 74

slides 3-4 have the definitions you need

http://www.math.tamu.edu/~shatalov/MATH%20151%20Project.pdf

Answer 75

OKay what about critical and inflection points?

Answer 76

critical points = points where the derivative is 0
inflection points = points where the second derivative is 0

Answer 77

Continuity?

Answer 78

there's a lot to know for this tbh

Answer 79

big thing: if a function is known to be differentiable on some interval it must be continuous

but if a function is known to be continuous that doesn't necessarily mean its differentiable

Answer 80

ex: f(x) = |x| is continuous but not differentiable at x = 0

Answer 81

and differentiable is just that a point exists right

Answer 82

differentiable = the derivative exists

Answer 83

okay and last thing ,, graphing funcation at f' and f''

Answer 84

any examples?

Answer 85

yes one sec

Answer 86

like, if they give you a graph f(x) and ask you to graph f'(x) you would graph the derivative

Answer 87

oh

well you have x(x^2-48)

to find the critical points, find where the derivative is 0
and to find the inflection points find where the second derivative is 0

Answer 88

f'(x) = 3x^2 - 48 = 0

so the critical points are x = -4 and x = 4

Answer 89

for the intervals where f'(x) is +, -, or 0, we have to set up the number line:

Answer 90

that tells you the function is increasing from -infinity to -4
decreasing from -4 to 4
increasing from 4 to infinity

Answer 91

same logic with the second derivative but instead of telling you where the function is increasing/decreasing, it tells you where the function is concave up or down

Answer 92

Got it

Answer 93

then once you have all the pieces, just put them together to make a function

Answer 94

IIs that all?

Answer 95

yeah