Question

find f ' (x) and f '(c) f(x)= 5x^-2 (x+3) value of c = 1

Answer 1

is your f(x) =\(5x^{-2}(x+3)\) ??

Answer 2

yep!

Answer 3

distribute 5x^-2
what is
5x^-2 * x = ?

Answer 4

is this us9ig the constant multiple?

Answer 5

*using

Answer 6

5x^-3?

Answer 7

x^m.x^n =x^{m+n}
so
x^-2 . x^1 = x^{-2+1} = ?

Answer 8

oh darn it's ^-1

Answer 9

yes, so u get 5x^-1
and 5*x^(-2)*3 = 15 x^(-2)
now do u know the differentiation of x^n ?

Answer 10

ok is this what i was supposed to do? distribute the (x+3)? if so i get..

Answer 11

no?
\(\huge\frac{d}{dx}x^n=nx^{n-1}\)

Answer 12

5x^-1 + 15x^-6 = 20x^-7

Answer 13

huh? no! that is not correct.
you cannot add exponents when you add terms, you add exponents, only when you multiply terms

Answer 14

ok, but was 5x^-1 + 15x^-6 correct?

Answer 15

and you have f(x) = 5x^(-1) +15x^(-2)

Answer 16

how -6 ?

Answer 17

oh darn. i made the same mistake again for the exponents.

Answer 18

f(x) = 5x^(-1) +15x^(-2)
now can u differentiate this using the formula i gave ?

Answer 19

so for 5x^-2 (3) part, i was only supposed to multiply the and 3? and leave the exponent alone? gotcha

Answer 20

*the 5 and 3

Answer 21

yes, because those are constants, x is variable, u leave x alone while multiplying constants

Answer 22

ok so using that formula i get this:

Answer 23

-5x^-2 + (-2)15x^-3

Answer 24

that is very correct!
just need simplification.
(-2)*15 = ?

Answer 25

-30. so then i add both parts together?

Answer 26

u get -5x^-2 -30x^-3, right ?
now u cannot simplify this further
because the exponents of x are different.

Answer 27

yes, i got those.

Answer 28

so that's it?

Answer 29

yes, that should be it
but you can also take -5x^(-2) common.

Answer 30

sorry, -5x^(-3) common to get -5x^(-3)(x+6)

Answer 31

where did the (x+6) come from?

Answer 32

-5x^-2 -30x^-3
=-5x^(-3) (x) - 6*5 x^(-3)
= -5x^(-3) (x+6)

Answer 33

ok so i get how you got -5x^-2 -30x^-3

Answer 34

when I look at =-5x^(-3) (x) - 6*5 x^(-3) i get confused

Answer 35

keep it as -5x^-2 -30x^-3 then .
now to find f'(c)
u just put x=c=1

Answer 36

ok?

Answer 37

so for f'c i get the same thing? since it's jsut 1

Answer 38

f'(x)= -5x^-2 -30x^-3 ,right ?
we have c=1.
so to get f'(1), u just put x=1 in f'(x)
that is , put x= 1 in -5x^-2 -30x^-3
make sense ?

Answer 39

in the back of my book it says:
f '(x)= 5[x^-2 (1) + (x+3)(-2x^-3)]
=-5(x+6) / x^3

Answer 40

so that's what you were trying to tell me i guess

Answer 41

yes.

Answer 42

and u still need tp put x=1

Answer 43

=-35

Answer 44

that is correct :)

Answer 45

:D