Question

derivatives
I cant seem to find the f' of these
P=.015v^3 v=13 mph
answer is 7.6w/mph
I tried .015(v+h)^3-.015(v)^3 but that doesnt seem to work, also
d=.05v^2+v Find R.O.C. when v=49mph
answer is 5.9
tried same process got 4.9 ?

Answer 1

first one f' is 3v^2=3(13)^2=507*.015=7.605
is that right?

Answer 2

0.045v^2 = dP/dv
Hence substituting we get 7.605
ROC would be dd/dv which is0.1v+1, hence 5.9
I dont understand what the problem is really :/

Answer 3

how did you get .1v+1

Answer 4

or how did you get .045v^2
you see how I am trying to work it

Answer 5

all i know if f(x+h)-f(x)/h

Answer 6

alright. when you differentiate, this happens
dy/dx (x^n) = nx^n-1
hence 3*0.015v^3-1 = 0.045v^2. Then substitute. Same goes for the other question.
Also, dy/dx(x) = 1
since x^1, 1*x (1-1) = x^0 = 1

Answer 7

so what ever th

Answer 8

so whatever the power is just multiply it to the coefficient -1?
not really following

Answer 9

do you not use
F(x+h)-f(x)/h ?

Answer 10

Multiply the power with the variable, and subtract one from the power. If you don't know this then i advise redoing your calculus because this is the very first thing they teach in acalculus class.
Sorry i dont know that formula i learnt calculus from a book and i did it by some formulas i saw

Answer 11

that formula is the first thing they taught me Im in calc 1 right now

Answer 12

thank you for the help

Answer 13

i will work on it

Answer 14

Ohh i assume you're at the very beginning. Apologies i don't know what the h stands for here. You'll be introduced to it real soon:)

Answer 15

i believe i understand your method much easier thank you

Answer 16

Haha you're welcome :) Yes its easier than dividing everything and splitting the function :)

Answer 17

true

Answer 18

quick q

Answer 19

on the second problem since x^0 is 1
.05(49)+1=3.45 not 5.9

Answer 20

nevermind

Answer 21

.1(49)+1=5.9

Answer 22

Haha its alright :) All teh best :)

Answer 23

thanks again

Answer 24

youre welcome :)