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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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