- anonymous

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

- jamiebookeater

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- anonymous

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

- anonymous

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

- myininaya

So what did you get for f'?

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## More answers

- myininaya

I will check it.

- anonymous

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

- myininaya

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

- myininaya

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

- anonymous

oh ok

- myininaya

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

- anonymous

so the awnser should be 19/2

- myininaya

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

- myininaya

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

- myininaya

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

- anonymous

So how does it apply

- myininaya

what do you mean?

- anonymous

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

- anonymous

correct?

- myininaya

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

- myininaya

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

- anonymous

Could you show me what you plugged in so I understand

- myininaya

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

- anonymous

No

- myininaya

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

- myininaya

(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).

- myininaya

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

- myininaya

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'

- anonymous

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

- myininaya

Yep

- myininaya

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

- anonymous

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

- anonymous

ok

- myininaya

h' is h'(x)

- myininaya

[-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)

- myininaya

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

- myininaya

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

- anonymous

Oh ok Now I get what you mean by this

- myininaya

I put the marks through the bad parts.

- myininaya

so you understand the constant multiple rule?

- anonymous

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

- myininaya

That is right.

- anonymous

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

- myininaya

Ok. Neatness.

- anonymous

Thank you so much

- myininaya

Np.

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