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

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

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

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

RohangrrBest 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_x3Best ResponseYou've already chosen the best response.1
derivative from the first principle..?
 one year ago

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

Mimi_x3Best 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_x3Best ResponseYou've already chosen the best response.1
mathslover; its from the first principle..
 one year ago

RohangrrBest 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

hartnnBest 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

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

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

RohangrrBest 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

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

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

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

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

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

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

CallistoBest 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

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

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

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