Question

Given that f(x)=-2h(x)+(2x/h(x)); h(4)=4, h'(4)=-2. Find f'(4)

Answer 1

well their is:
a)13/2
b)7/2
c)-5/2
d)11/2
e)17/2

Answer 2

I tried solving for the derivative of f(x) but I must be doing something wrong or the awnser is not there

Answer 3

So what did you get for f'?

Answer 4

I will check it.

Answer 5

\[f'(x) = h(x) + -2h'(x)+ (\frac{ 2h(x)-(2x)h(x) }{ h(x)^2 }\]

Answer 6

Also the answer is there.
h(x) shouldn't be there.
you are missing h' in your quotient

Answer 7

(-2h)'=-2(h)'=-2h'

Answer 8

oh ok

Answer 9

(2x/h)'=[(2x)'h-2x(h)']/h^2

Answer 10

so the awnser should be 19/2

Answer 11

hmm... that isn't what I got...let me check my result.

Answer 12

Oh you are still assuming h is there aren't you?

Answer 13

(-2h)'=-2(h)'=-2h' not h-2h'

Answer 14

So how does it apply

Answer 15

what do you mean?

Answer 16

well I am plugging in everything where
\[h(4)+(-2h'(4)) + \frac{ (2)(h(4))-(2x)(h'(4)) }{ h(4)^2 } = 4 +(-2(-2)) + \frac{ (2)(4)-(2(4))(-2) }{4^2 }= f'(4)\]

Answer 17

correct?

Answer 18

Well see I keep telling you (-2h)'=-2h' not h-2h'

Answer 19

That first term you are putting in shouldn't be there.

Answer 20

Could you show me what you plugged in so I understand

Answer 21

So you don't understand why (-2h)' is -2h' and not h-2h'?

Answer 22

No

Answer 23

I'm liking I'm using the constant multiple rule.

Answer 24

(cf)'=cf'
Let k(x)=cf(x)
To find k'(x) we use the formal definition of derivative:
\[k'(x)=\lim_{\Delta x \rightarrow 0} \frac{k(x+\Delta x)-k(x)}{\Delta x}=\lim_{\Delta x \rightarrow 0} \frac{cf(x+\Delta x)-cf(x)}{\Delta x}\]
Note: By the distributive property ab+ac=a(b+c)
\[=\lim_{\Delta x \rightarrow 0} c \frac{f(x+\Delta x)-f(x)}{\Delta x} \]
Note: By limit properties, we can do lim c g(x) = c lim g(x)
\[=c \lim_{\Delta x \rightarrow 0} \frac{f(x+\Delta x)-f(x)}{\Delta x} \]
Note: By definition of derivative that above is cf'(x).

Answer 25

The constant multiple rule:
k=cf
k'=cf'

Answer 26

So say k=ch
then k'=ch'

That c means constant
The constant in front of h in your problem is -2
so k=-2h
then k'=-2h'

Answer 27

So when I plug in for -2h' when h(4)=4 and h'(4)=-2

Answer 28

Yep

Answer 29

-2h'(x)
x is 4.
-2h'(4)
-2h'(4)=-2(-2)=4
Now simplify your quotient and add.

Answer 30

Does this mean -2h'= -2(-2) or -2(4)

Answer 31

ok

Answer 32

h' is h'(x)

Answer 33

[-2h(x)] ' =-2 [ h(x)] ' =-2 h'(x)
This has been what I have been saying but you wrote it as h(x)-2h'(x)

Answer 34

\[\cancel{h(4)}+(-2h'(4)) + \frac{ (2)(h(4))-(2x)(h'(4)) }{ h(4)^2 } = \cancel{4} +(-2(-2)) + \frac{ (2)(4)-(2(4))(-2) }{4^2 }= f'(4)\]

Answer 35

This is what you wrote earlier
I'm telling you (-2h)'=-2h' not h-2h'
do you understand now?

Answer 36

Oh ok Now I get what you mean by this

Answer 37

I put the marks through the bad parts.

Answer 38

so you understand the constant multiple rule?

Answer 39

Yes I do now understand the multiple rule and the awnser is 11/2

Answer 40

That is right.

Answer 41

I just didnt understand it conceptually but now that I see it plugged in and used I do now

Answer 42

Ok. Neatness.

Answer 43

Thank you so much

Answer 44

Np.