Question

Suppose R'(1)=16.
a) If f(x)=R(5x), find f '(1/5)
b) If g(x)=R(x+8), find g'(−7)
c)If h(x) = R(x/7), find h'(7)

Answer 1

@iambatman can you help me please?

Answer 2

If f(x)=R(5x), then f'(x) = ??

Answer 3

Right, I don't know if I am supposed to use the chain rule, but if so, what is the inside and what is the outside?

Answer 4

the outer function is R(x)

the inner function is 5x

Answer 5

ok,so would it be R'(5x)*5? which would be 16*5=80

Answer 6

because 5(1/5)=1 and R'(1)=16

Answer 7

not quite

Answer 8

oh wait, nvm

Answer 9

f(x)=R(5x)

f'(x)=5*R'(5x)

f'(1/5)=5*R'(5(1/5))

f'(1/5)=5*R'(1)

f'(1/5)=5*16

f'(1/5)=80

yeah you got it right

Answer 10

ok, and so for b), would the answer be -112 because after it is all simplified, you get 16*-7

Answer 11

g(x)=R(x+8)

g'(x)=R'(x+8)

g'(-7)=R'(-7+8)

g'(-7)=R'(1)

g'(-7)=16

so this shows you that shifting the graph (or translating it) does not affect the slopes of the tangent lines.

Answer 12

I didn't show it, but with the chain rule you'll have 1* on the outside, but that just results in R'(x+8)

Answer 13

ok, I see what I did wrong. I put it as R'(-7+8)*(x) which gave me R'(1)=16*-7. however the derivative of x is 1 which I messed up.

Answer 14

I go the answer for c) by doing the chain and quotient rule. the answer I got was 16/7

Answer 15

no you would derive x+8 to get 1 and then multiply it by R'(x+8)

Answer 16

h(x) = R(x/7)

h'(x) = (1/7)*R'(x/7)

h'(7) = (1/7)*R'(7/7)

h'(7) = (1/7)*R'(1)

h'(7) = (1/7)*16

h'(7) = 16/7

looks good

Answer 17

you don't need to do the quotient rule, but it's possible to use it

Answer 18

I'd just pull out the coefficient

Answer 19

ok, thanks!! I just have one last question to do for my homework. Can you also help me with that really quick?

Answer 20

sure

Answer 21

I need to find the derivative of g(x)=5/(a^2-w^2)^4

My answer was -40a(a^2-w^2)^3/(a^2-w^2)^8

Answer 22

\[\frac{ -40a(a^2-w^2)^3 }{ (a^2-w^2)^8 }\] to make my answer more clear.

Answer 23

I see g(x), but I don't see any x on the right side

Answer 24

sorry I meant to put g(w) but habit forced me to put x

Answer 25

how did you get -40a? When you derive the inside a^2 - w^2, you should be left with 2w (the a^2 is constant which derives to 0)

Answer 26

-2w I mean

Answer 27

now that you mention that, I have no clue because I got -40a by -5*8a and I got 8a by 4*2a

Answer 28

it's very close to the answer, but incorrect

Answer 29

you can't say a^2 turns into 2a because you aren't deriving with respect to 'a'

Answer 30

right, I always get confused when there are just letters...so would the new answer be -10w(a^2-w^2)^3/(a^2-w^2)^8

Answer 31

the general rule is to derive with respect to the independent variable

Answer 32

usually (not always), the other letters are constants

Answer 33

right, I get that, it just always trips me up...

Answer 34

-10w(a^2-w^2)^3/(a^2-w^2)^8 is close, but again, incorrect

Answer 35

what part of it is incorrect?

Answer 36

it might help to think of the problem in the form u/v

Answer 37

the -10w, the (a^2-w^2)^3, or the bottom?

Answer 38

what do you mean u/v?

Answer 39

u = 5
v = (a^2-w^2)^4

Answer 40

so

u' = 0
v' = 4(a^2-w^2)*2w = 8w(a^2-w^2)

Answer 41

this is setting up for the quotient rule

Answer 42

oops, v' = 4(a^2-w^2)*(-2w) = -8w(a^2-w^2), there fixed it

Answer 43

ok, so it would be -40w.....

Answer 44

first off, is the bottom part of my answer correct?

Answer 45

oh and I completely forgot an exponent

v' = 4(a^2-w^2)^3*(-2w) = -8w(a^2-w^2)^3

Answer 46

or wait is it just 40w since it is -5*-8w...

Answer 47

anyways,

u'*v - v'*u = 0*(a^2-w^2)^4 - ( -8w(a^2-w^2)^3 )*5 = 40w(a^2-w^2)^3

is all in the numerator

Answer 48

ok, thanks!! I think it have it now. \[\frac{ 40w(a^2-w^2)^3 }{ (a^2-w^2)^8 }\]

Answer 49

then you can reduce to get

\[\large \frac{ 40w(a^2-w^2)^3 }{ (a^2-w^2)^8 } = \frac{40w}{(a^2-w^2)^5 } \]

Answer 50

ok, thanks!!! I finally get it!!

Answer 51

np