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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

myininaya
 one year ago
Best 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).

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

myininaya
 one year ago
Best ResponseYou've already chosen the best response.3So 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'

Matthew071
 one year ago
Best ResponseYou've already chosen the best response.0So when I plug in for 2h' when h(4)=4 and h'(4)=2

myininaya
 one year ago
Best ResponseYou've already chosen the best response.32h'(x) x is 4. 2h'(4) 2h'(4)=2(2)=4 Now simplify your quotient and add.

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

myininaya
 one year ago
Best 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)

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

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

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

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

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

Matthew071
 one year ago
Best ResponseYou've already chosen the best response.0Yes I do now understand the multiple rule and the awnser is 11/2

Matthew071
 one year ago
Best ResponseYou've already chosen the best response.0I just didnt understand it conceptually but now that I see it plugged in and used I do now
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