anonymous
  • anonymous
Given that f(x)=-2h(x)+(2x/h(x)); h(4)=4, h'(4)=-2. Find f'(4)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
well their is: a)13/2 b)7/2 c)-5/2 d)11/2 e)17/2
anonymous
  • anonymous
I tried solving for the derivative of f(x) but I must be doing something wrong or the awnser is not there
myininaya
  • myininaya
So what did you get for f'?

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myininaya
  • myininaya
I will check it.
anonymous
  • anonymous
\[f'(x) = h(x) + -2h'(x)+ (\frac{ 2h(x)-(2x)h(x) }{ h(x)^2 }\]
myininaya
  • myininaya
Also the answer is there. h(x) shouldn't be there. you are missing h' in your quotient
myininaya
  • myininaya
(-2h)'=-2(h)'=-2h'
anonymous
  • anonymous
oh ok
myininaya
  • myininaya
(2x/h)'=[(2x)'h-2x(h)']/h^2
anonymous
  • anonymous
so the awnser should be 19/2
myininaya
  • myininaya
hmm... that isn't what I got...let me check my result.
myininaya
  • myininaya
Oh you are still assuming h is there aren't you?
myininaya
  • myininaya
(-2h)'=-2(h)'=-2h' not h-2h'
anonymous
  • anonymous
So how does it apply
myininaya
  • myininaya
what do you mean?
anonymous
  • anonymous
well I am plugging in everything where \[h(4)+(-2h'(4)) + \frac{ (2)(h(4))-(2x)(h'(4)) }{ h(4)^2 } = 4 +(-2(-2)) + \frac{ (2)(4)-(2(4))(-2) }{4^2 }= f'(4)\]
anonymous
  • anonymous
correct?
myininaya
  • myininaya
Well see I keep telling you (-2h)'=-2h' not h-2h'
myininaya
  • myininaya
That first term you are putting in shouldn't be there.
anonymous
  • anonymous
Could you show me what you plugged in so I understand
myininaya
  • myininaya
So you don't understand why (-2h)' is -2h' and not h-2h'?
anonymous
  • anonymous
No
myininaya
  • myininaya
I'm liking I'm using the constant multiple rule.
myininaya
  • myininaya
(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
  • myininaya
The constant multiple rule: k=cf k'=cf'
myininaya
  • myininaya
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'
anonymous
  • anonymous
So when I plug in for -2h' when h(4)=4 and h'(4)=-2
myininaya
  • myininaya
Yep
myininaya
  • myininaya
-2h'(x) x is 4. -2h'(4) -2h'(4)=-2(-2)=4 Now simplify your quotient and add.
anonymous
  • anonymous
Does this mean -2h'= -2(-2) or -2(4)
anonymous
  • anonymous
ok
myininaya
  • myininaya
h' is h'(x)
myininaya
  • myininaya
[-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
  • myininaya
\[\cancel{h(4)}+(-2h'(4)) + \frac{ (2)(h(4))-(2x)(h'(4)) }{ h(4)^2 } = \cancel{4} +(-2(-2)) + \frac{ (2)(4)-(2(4))(-2) }{4^2 }= f'(4)\]
myininaya
  • myininaya
This is what you wrote earlier I'm telling you (-2h)'=-2h' not h-2h' do you understand now?
anonymous
  • anonymous
Oh ok Now I get what you mean by this
myininaya
  • myininaya
I put the marks through the bad parts.
myininaya
  • myininaya
so you understand the constant multiple rule?
anonymous
  • anonymous
Yes I do now understand the multiple rule and the awnser is 11/2
myininaya
  • myininaya
That is right.
anonymous
  • anonymous
I just didnt understand it conceptually but now that I see it plugged in and used I do now
myininaya
  • myininaya
Ok. Neatness.
anonymous
  • anonymous
Thank you so much
myininaya
  • myininaya
Np.

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