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Rohangrr
Group Title
Use the limit definition to compute the derivative, f'(x),
\[f(x)=\frac { 1 }{ 2 } x\frac { 3 }{ 5 } \]
 one year ago
 one year ago
Rohangrr Group Title
Use the limit definition to compute the derivative, f'(x), \[f(x)=\frac { 1 }{ 2 } x\frac { 3 }{ 5 } \]
 one year ago
 one year ago

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mathslover Group TitleBest ResponseYou've already chosen the best response.0
oh k progression or in progress? MAN AT WORK :P
 one year ago

chemENGINEER Group TitleBest ResponseYou've already chosen the best response.0
Answer on Progress
 one year ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.0
Oh wow what a question...!!
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
rohangrr is typing a reply...
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
oh k here we go
 one year ago

Rohangrr Group TitleBest ResponseYou've already chosen the best response.4
Use the limit definition to compute the derivative, f'(x), \[f(x)=\frac { 1 }{ 2 } x\frac { 3 }{ 5 } \]
 one year ago

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.1
derivative from the first principle..?
 one year ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.0
I am not typing..
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
dw:1345962336537:dw
 one year ago

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.1
\[\frac{dy}{dx} = \lim_{x \rightarrow 0} \frac{f(x+h)f(x)}{h} \]
 one year ago

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.1
mathslover; its from the first principle..
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
oh sorry
 one year ago

Rohangrr Group TitleBest ResponseYou've already chosen the best response.4
\[\huge f'\quad (x)\quad =\lim _{ x\rightarrow 0 } \frac { f(x+\triangle x)f(x) }{ \triangle x } \] \[\huge \quad =\lim _{ x\rightarrow 0 } \frac { \frac { 1 }{ 2 } (x+\triangle x)\frac { 3 }{ 5 } \{ \frac { 1 }{ 2 } x\frac { 3 }{ 5 } \} }{ \triangle x } \] What will be after this
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
first of all the limit would be h>0 !! then u cancel,x/2 and 3/5 in numerator.....
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
i mean delta x >0
 one year ago

Rohangrr Group TitleBest ResponseYou've already chosen the best response.4
Okay then
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
then u will simply get:dw:1345962738189:dw
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
the same I got but my method was wrong :(
 one year ago

Rohangrr Group TitleBest ResponseYou've already chosen the best response.4
\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 }
 one year ago

Rohangrr Group TitleBest ResponseYou've already chosen the best response.4
is this the final answer
 one year ago

Rohangrr Group TitleBest ResponseYou've already chosen the best response.4
dw:1345963107126:dw
 one year ago

Rohangrr Group TitleBest ResponseYou've already chosen the best response.4
Srously IM confused
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
dw:1345963124504:dw
 one year ago

Rohangrr Group TitleBest ResponseYou've already chosen the best response.4
dw:1345963183321:dw
 one year ago

Rohangrr Group TitleBest ResponseYou've already chosen the best response.4
dw:1345963244684:dw
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.0
Do you still have a question?
 one year ago

Rohangrr Group TitleBest ResponseYou've already chosen the best response.4
Hi @myininaya I didnt observe u here
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.0
Yes. I'm here.
 one year ago

Rohangrr Group TitleBest ResponseYou've already chosen the best response.4
Thanks guys
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.3
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!
 one year ago

Rohangrr Group TitleBest ResponseYou've already chosen the best response.4
But u r still alive
 one year ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.0
But I am still in confusion now..
 one year ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.0
To Whom should I give the medal ??
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
what confusion u have?
 one year ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.0
No, it is now clear...
 one year ago
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