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Matthew071
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Given that f(x)=2h(x)+(2x/h(x)); h(4)=4, h'(4)=2. Find f'(4)
 10 months ago
 10 months ago
Matthew071 Group Title
Given that f(x)=2h(x)+(2x/h(x)); h(4)=4, h'(4)=2. Find f'(4)
 10 months ago
 10 months ago

This Question is Closed

Matthew071 Group TitleBest ResponseYou've already chosen the best response.0
well their is: a)13/2 b)7/2 c)5/2 d)11/2 e)17/2
 10 months ago

Matthew071 Group TitleBest ResponseYou've already chosen the best response.0
I tried solving for the derivative of f(x) but I must be doing something wrong or the awnser is not there
 10 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.3
So what did you get for f'?
 10 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.3
I will check it.
 10 months ago

Matthew071 Group TitleBest ResponseYou've already chosen the best response.0
\[f'(x) = h(x) + 2h'(x)+ (\frac{ 2h(x)(2x)h(x) }{ h(x)^2 }\]
 10 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.3
Also the answer is there. h(x) shouldn't be there. you are missing h' in your quotient
 10 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.3
(2h)'=2(h)'=2h'
 10 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.3
(2x/h)'=[(2x)'h2x(h)']/h^2
 10 months ago

Matthew071 Group TitleBest ResponseYou've already chosen the best response.0
so the awnser should be 19/2
 10 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.3
hmm... that isn't what I got...let me check my result.
 10 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.3
Oh you are still assuming h is there aren't you?
 10 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.3
(2h)'=2(h)'=2h' not h2h'
 10 months ago

Matthew071 Group TitleBest ResponseYou've already chosen the best response.0
So how does it apply
 10 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.3
what do you mean?
 10 months ago

Matthew071 Group TitleBest ResponseYou've already chosen the best response.0
well I am plugging in everything where \[h(4)+(2h'(4)) + \frac{ (2)(h(4))(2x)(h'(4)) }{ h(4)^2 } = 4 +(2(2)) + \frac{ (2)(4)(2(4))(2) }{4^2 }= f'(4)\]
 10 months ago

Matthew071 Group TitleBest ResponseYou've already chosen the best response.0
correct?
 10 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.3
Well see I keep telling you (2h)'=2h' not h2h'
 10 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.3
That first term you are putting in shouldn't be there.
 10 months ago

Matthew071 Group TitleBest ResponseYou've already chosen the best response.0
Could you show me what you plugged in so I understand
 10 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.3
So you don't understand why (2h)' is 2h' and not h2h'?
 10 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.3
I'm liking I'm using the constant multiple rule.
 10 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.3
(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).
 10 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.3
The constant multiple rule: k=cf k'=cf'
 10 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.3
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'
 10 months ago

Matthew071 Group TitleBest ResponseYou've already chosen the best response.0
So when I plug in for 2h' when h(4)=4 and h'(4)=2
 10 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.3
2h'(x) x is 4. 2h'(4) 2h'(4)=2(2)=4 Now simplify your quotient and add.
 10 months ago

Matthew071 Group TitleBest ResponseYou've already chosen the best response.0
Does this mean 2h'= 2(2) or 2(4)
 10 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.3
h' is h'(x)
 10 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.3
[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)
 10 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.3
\[\cancel{h(4)}+(2h'(4)) + \frac{ (2)(h(4))(2x)(h'(4)) }{ h(4)^2 } = \cancel{4} +(2(2)) + \frac{ (2)(4)(2(4))(2) }{4^2 }= f'(4)\]
 10 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.3
This is what you wrote earlier I'm telling you (2h)'=2h' not h2h' do you understand now?
 10 months ago

Matthew071 Group TitleBest ResponseYou've already chosen the best response.0
Oh ok Now I get what you mean by this
 10 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.3
I put the marks through the bad parts.
 10 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.3
so you understand the constant multiple rule?
 10 months ago

Matthew071 Group TitleBest ResponseYou've already chosen the best response.0
Yes I do now understand the multiple rule and the awnser is 11/2
 10 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.3
That is right.
 10 months ago

Matthew071 Group TitleBest ResponseYou've already chosen the best response.0
I just didnt understand it conceptually but now that I see it plugged in and used I do now
 10 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.3
Ok. Neatness.
 10 months ago

Matthew071 Group TitleBest ResponseYou've already chosen the best response.0
Thank you so much
 10 months ago
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