anonymous
  • anonymous
Use the limit definition to compute the derivative, f'(x), \[f(x)=\frac { 1 }{ 2 } x-\frac { 3 }{ 5 } \]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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mathslover
  • mathslover
oh k progression or in progress? MAN AT WORK :P
anonymous
  • anonymous
Answer on Progress
anonymous
  • anonymous
Oh wow what a question...!!

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More answers

mathslover
  • mathslover
rohangrr is typing a reply...
mathslover
  • mathslover
oh k here we go
anonymous
  • anonymous
Use the limit definition to compute the derivative, f'(x), \[f(x)=\frac { 1 }{ 2 } x-\frac { 3 }{ 5 } \]
Mimi_x3
  • Mimi_x3
derivative from the first principle..?
anonymous
  • anonymous
I am not typing..
mathslover
  • mathslover
|dw:1345962336537:dw|
Mimi_x3
  • Mimi_x3
\[\frac{dy}{dx} = \lim_{x \rightarrow 0} \frac{f(x+h)-f(x)}{h} \]
Mimi_x3
  • Mimi_x3
mathslover; its from the first principle..
mathslover
  • mathslover
oh sorry
anonymous
  • anonymous
\[\huge f'\quad (x)\quad =\lim _{ x\rightarrow 0 } \frac { f(x+\triangle x)-f(x) }{ \triangle x } \] \[\huge \quad =\lim _{ x\rightarrow 0 } \frac { \frac { 1 }{ 2 } (x+\triangle x)-\frac { 3 }{ 5 } -\{ \frac { 1 }{ 2 } x-\frac { 3 }{ 5 } \} }{ \triangle x } \] What will be after this
hartnn
  • hartnn
first of all the limit would be h->0 !! then u cancel,x/2 and 3/5 in numerator.....
hartnn
  • hartnn
i mean delta x ->0
anonymous
  • anonymous
Okay then
hartnn
  • hartnn
then u will simply get:|dw:1345962738189:dw|
mathslover
  • mathslover
the same I got but my method was wrong :(
anonymous
  • anonymous
\quad =\lim _{ \triangle x\rightarrow 0 } \frac { \frac { 1 }{ 2 } x+\frac { 1 }{ 2 } \triangle x-\frac { 3 }{ 5 } -\frac { 1 }{ 2 } x+\frac { 3 }{ 5 } \} }{ \triangle x }
anonymous
  • anonymous
is this the final answer
anonymous
  • anonymous
|dw:1345963107126:dw|
anonymous
  • anonymous
Srously IM confused
hartnn
  • hartnn
|dw:1345963124504:dw|
anonymous
  • anonymous
|dw:1345963183321:dw|
anonymous
  • anonymous
|dw:1345963244684:dw|
myininaya
  • myininaya
Do you still have a question?
anonymous
  • anonymous
Hi @myininaya I didnt observe u here
myininaya
  • myininaya
Yes. I'm here.
anonymous
  • anonymous
Thanks guys
Callisto
  • Callisto
First principle: \[f'(x) =\lim_{Δ x \rightarrow 0}\frac{f(x+Δ x)-f(x)}{Δ x}\] When \(f(x) = \frac{1}{2}x - \frac{3}{5}\), \[f'(x) =\lim_{Δ x \rightarrow 0}\frac{[\frac{1}{2}(x+Δ x)-\frac{3}{5}]-(\frac{1}{2}x - \frac{3}{5})}{Δ x}\] \[=\lim_{Δ x \rightarrow 0}\frac{\frac{1}{2}x+\frac{1}{2}Δ x-\frac{3}{5}-\frac{1}{2}x + \frac{3}{5}}{Δ x}\] \[=\lim_{Δ x \rightarrow 0}\frac{\frac{1}{2}Δ x}{Δ x}\] \[=\lim_{Δ x \rightarrow 0}\frac{\frac{1}{2}\cancel{Δ x}}{\cancel{Δ x}}\] \[=\lim_{Δ x \rightarrow 0}\frac{1}{2}\] \[=...\] Wow! I'm tooooo late!
anonymous
  • anonymous
But u r still alive
anonymous
  • anonymous
But I am still in confusion now..
anonymous
  • anonymous
To Whom should I give the medal ??
hartnn
  • hartnn
what confusion u have?
anonymous
  • anonymous
No, it is now clear...
hartnn
  • hartnn
ok..

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