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

1. mathslover

oh k progression or in progress? MAN AT WORK :P

2. chemENGINEER

3. waterineyes

Oh wow what a question...!!

4. mathslover

5. mathslover

oh k here we go

6. Rohangrr

Use the limit definition to compute the derivative, f'(x), $f(x)=\frac { 1 }{ 2 } x-\frac { 3 }{ 5 }$

7. Mimi_x3

derivative from the first principle..?

8. waterineyes

I am not typing..

9. mathslover

|dw:1345962336537:dw|

10. Mimi_x3

$\frac{dy}{dx} = \lim_{x \rightarrow 0} \frac{f(x+h)-f(x)}{h}$

11. Mimi_x3

mathslover; its from the first principle..

12. mathslover

oh sorry

13. Rohangrr

$\huge f'\quad (x)\quad =\lim _{ x\rightarrow 0 } \frac { f(x+\triangle x)-f(x) }{ \triangle x }$ $\huge \quad =\lim _{ x\rightarrow 0 } \frac { \frac { 1 }{ 2 } (x+\triangle x)-\frac { 3 }{ 5 } -\{ \frac { 1 }{ 2 } x-\frac { 3 }{ 5 } \} }{ \triangle x }$ What will be after this

14. hartnn

first of all the limit would be h->0 !! then u cancel,x/2 and 3/5 in numerator.....

15. hartnn

i mean delta x ->0

16. Rohangrr

Okay then

17. hartnn

then u will simply get:|dw:1345962738189:dw|

18. mathslover

the same I got but my method was wrong :(

19. Rohangrr

\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 }

20. Rohangrr

21. Rohangrr

|dw:1345963107126:dw|

22. Rohangrr

Srously IM confused

23. hartnn

|dw:1345963124504:dw|

24. Rohangrr

|dw:1345963183321:dw|

25. Rohangrr

|dw:1345963244684:dw|

26. myininaya

Do you still have a question?

27. Rohangrr

Hi @myininaya I didnt observe u here

28. myininaya

Yes. I'm here.

29. Rohangrr

Thanks guys

30. 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!

31. Rohangrr

But u r still alive

32. waterineyes

But I am still in confusion now..

33. waterineyes

To Whom should I give the medal ??

34. hartnn

what confusion u have?

35. waterineyes

No, it is now clear...

36. hartnn

ok..