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

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Matthew071
 2 years 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
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.3So what did you get for f'?

Matthew071
 2 years 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
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.3(2x/h)'=[(2x)'h2x(h)']/h^2

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

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

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

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

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

Matthew071
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.3Well see I keep telling you (2h)'=2h' not h2h'

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

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

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

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

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

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

myininaya
 2 years 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
 2 years 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
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0Oh ok Now I get what you mean by this

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

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

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

Matthew071
 2 years 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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