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

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mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0oh k progression or in progress? MAN AT WORK :P

chemENGINEER
 3 years ago
Best ResponseYou've already chosen the best response.0Answer on Progress

waterineyes
 3 years ago
Best ResponseYou've already chosen the best response.0Oh wow what a question...!!

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0rohangrr is typing a reply...

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

Mimi_x3
 3 years ago
Best ResponseYou've already chosen the best response.1derivative from the first principle..?

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1345962336537:dw

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

Mimi_x3
 3 years ago
Best ResponseYou've already chosen the best response.1mathslover; its from the first principle..

Rohangrr
 3 years ago
Best 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

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

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.3then u will simply get:dw:1345962738189:dw

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0the same I got but my method was wrong :(

Rohangrr
 3 years ago
Best 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 }

Rohangrr
 3 years ago
Best ResponseYou've already chosen the best response.4is this the final answer

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.0Do you still have a question?

Rohangrr
 3 years ago
Best ResponseYou've already chosen the best response.4Hi @myininaya I didnt observe u here

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.3First 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!

waterineyes
 3 years ago
Best ResponseYou've already chosen the best response.0But I am still in confusion now..

waterineyes
 3 years ago
Best ResponseYou've already chosen the best response.0To Whom should I give the medal ??

waterineyes
 3 years ago
Best ResponseYou've already chosen the best response.0No, it is now clear...
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