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Matthew071Best 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
 6 months ago

Matthew071Best 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
 6 months ago

myininayaBest ResponseYou've already chosen the best response.3
So what did you get for f'?
 6 months ago

Matthew071Best ResponseYou've already chosen the best response.0
\[f'(x) = h(x) + 2h'(x)+ (\frac{ 2h(x)(2x)h(x) }{ h(x)^2 }\]
 6 months ago

myininayaBest 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
 6 months ago

myininayaBest ResponseYou've already chosen the best response.3
(2x/h)'=[(2x)'h2x(h)']/h^2
 6 months ago

Matthew071Best ResponseYou've already chosen the best response.0
so the awnser should be 19/2
 6 months ago

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

myininayaBest ResponseYou've already chosen the best response.3
Oh you are still assuming h is there aren't you?
 6 months ago

myininayaBest ResponseYou've already chosen the best response.3
(2h)'=2(h)'=2h' not h2h'
 6 months ago

Matthew071Best ResponseYou've already chosen the best response.0
So how does it apply
 6 months ago

Matthew071Best ResponseYou've already chosen the best response.0
well I am plugging in everything where \[h(4)+(2h'(4)) + \frac{ (2)(h(4))(2x)(h'(4)) }{ h(4)^2 } = 4 +(2(2)) + \frac{ (2)(4)(2(4))(2) }{4^2 }= f'(4)\]
 6 months ago

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

myininayaBest ResponseYou've already chosen the best response.3
That first term you are putting in shouldn't be there.
 6 months ago

Matthew071Best ResponseYou've already chosen the best response.0
Could you show me what you plugged in so I understand
 6 months ago

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

myininayaBest ResponseYou've already chosen the best response.3
I'm liking I'm using the constant multiple rule.
 6 months ago

myininayaBest 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).
 6 months ago

myininayaBest ResponseYou've already chosen the best response.3
The constant multiple rule: k=cf k'=cf'
 6 months ago

myininayaBest 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'
 6 months ago

Matthew071Best ResponseYou've already chosen the best response.0
So when I plug in for 2h' when h(4)=4 and h'(4)=2
 6 months ago

myininayaBest 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.
 6 months ago

Matthew071Best ResponseYou've already chosen the best response.0
Does this mean 2h'= 2(2) or 2(4)
 6 months ago

myininayaBest 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)
 6 months ago

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

myininayaBest 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?
 6 months ago

Matthew071Best ResponseYou've already chosen the best response.0
Oh ok Now I get what you mean by this
 6 months ago

myininayaBest ResponseYou've already chosen the best response.3
I put the marks through the bad parts.
 6 months ago

myininayaBest ResponseYou've already chosen the best response.3
so you understand the constant multiple rule?
 6 months ago

Matthew071Best ResponseYou've already chosen the best response.0
Yes I do now understand the multiple rule and the awnser is 11/2
 6 months ago

Matthew071Best 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
 6 months ago
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