Question

Using first principle, how do you find f'(x) if f(x) = 3/2x^2?

Answer 1

What IS the first principle? :)

Answer 2

though, I'm suspecting it's this...
\[\huge \frac{d}{dx} ax^n=nax^{n-1}\]

Answer 3

in first principle, f(x) is taken as (fx+h) and h is limited to zero.

f(x+h)-f(x) lim h->0

Answer 4

Yeah, i don't get calculus :p

Answer 5

It's solved by L'hospital's law...

Answer 6

I don't remember learning that during lesson :/

Answer 7

Okay, we'll solve it, step by step.

As the first principle states it.

\[\lim_{h \rightarrow 0} \frac{ f(x+h) -f(x)}{ h }\]

Answer 8

The equations seems familiar :) Thanks for taking time off to help me by the way, really appreciate it :)

Answer 9

By L'hospital's law, we have, \[\lim_{h \rightarrow 0} f'(x+h) - f'(x)\]

Answer 10

You can put the equation, key is, it is f'(x) wrt h.

Answer 11

Don't you just replace f(x) with the 3/2x^2? Cause then you wouldn't have to use the L'hospital's law?

Answer 12

Wait, I effed up there^

Answer 13

\[\lim_{h \rightarrow 0} \frac{ 3(x+h) ^{2} - 3(x)^{2}/2 }{ h }\]

Answer 14

do the 3(x)square at the end is divided by 2?

Answer 15

does*

Answer 16

\[\lim_{h \rightarrow 0} \frac{ 3(2xh + h^{2} }{ 2h }\]

Answer 17

f(x) equals \[\frac{ 3 }{ 2x ^{2} }\] by the way?

Answer 18

NOW you apply that L'hospital's law.

Answer 19

Oh... okay, will solve it that way. =P Sorry, this eqn editor is somewhat confusing.

Answer 20

Aha, yeah :p thank you for helping me by the way :)

Answer 21

Both of you, or whoever that helps me on here :p *

Answer 22

Using L'Hopital with the definition of the derivative leads to a tautology.

Answer 23

It's to solve by the first principle.... So, I guess that ought to be right.

Answer 24

\[\Large f'(x) =\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}\]
\[\Large = \lim_{h\rightarrow 0}\frac{\frac{d}{dh}[f(x+h)-f(x)]}{\frac{d}{dh}(h)}\]
\[\Large \lim_{h\rightarrow 0}f'(x+h)=f'(x)\]

So in the end, using L'Hopital with the definition of the derivative just leads to
\[\Large f'(x)=f'(x)\]

Answer 25

^thankyou. =) That's what I was trying to prove. =)

Answer 26

Yeah, but still, you'll have to differentiate 3/2(x^2) when that's what you were supposed to be finding! :(

Answer 27

But how am i suppose to do it using the first principle? :/

Answer 28

IKR, that's why first principle sucks.

Answer 29

Imagine doing that
\[\huge \lim_{h\rightarrow 0}\frac{\frac{3}{2(x+h)^2}-\frac{3}{2x^2}}{h}\]

Answer 30

And you can't use L'Hopital without already knowing beforehand what
\[\huge \frac{d}{dx}\frac{3}{2x^2}\]is.

Answer 31

Especially when i've never heard of the L'Hospital thingy that you guys/girls were talking about before :p

Answer 32

^WHICH is what makes First principle stupid.

Answer 33

I didn't expect you to :)
You only get to L'Hopital after a certain mastery of derivatives.
The first principle is just the definition, maybe to make you better appreciate the simplicity of the rules regarding derivatives ^.6

Answer 34

SO, Do i get this or have i done something wrong in the process? :p\[\lim_{h \rightarrow 0} \frac{ 6x ^{2}-3(2x ^{2}+h) }{ h2x ^{2}(2x ^{2}+h) }\]

Answer 35

A little off with the exponents
\[\Large \lim_{h \rightarrow 0} \frac{ 6x ^{2}-3\cdot2(x +h)^2 }{ h\cdot2x ^{2}\cdot2(x +h)^2 }\]

Answer 36

You know what, let's do it from the top :)
\[\huge \frac{\frac{3}{2(x+h)^2}-\frac{3}{2x^2}}{h}\]

Answer 37

That would be lovely thank you :)

Answer 38

Cross multiplying leads to
\[\huge \frac{\frac{6x^2-6(x+h)^2}{4x^2(x+h)^2}}{h}\]

Answer 39

So many fractions :p

Answer 40

And make it just one fraction
\[\huge \frac{6x^2-6(x+h)^2}{h\cdot 4x^2(x+h)^2}\]

Answer 41

Well? Don't be shy, evaluate ^.^
\[\huge \frac{6x^2-6(x^2+2xh+h^2)}{h\cdot 4x^2(x^2+2xh+h^2)}\]

Answer 42

So apparently now there's suppose to be a cancelling of the original divisor at this step?

Answer 43

h in the denominator should cancel. Which it will.

Answer 44

at this step or in a later step? :p

Answer 45

Almost there. Simplify the numerator.

Answer 46

-12xh - 6hsquare?

Answer 47

Yeah...
\[\huge \frac{h(-12x-h)}{h\cdot 4x^2(x^2+2xh+h^2)}\]
Now you can cancel out one h, which is more than enough, and you'll be able to get the limit as h goes to zero :)

Answer 48

and then expand the bottom bit?

Answer 49

expand, or not, cancel out the ones with h's
remember, h goes to ZERO.

Answer 50

But cancel out first
\[\huge \frac{\cancel h(-12x-h)}{\cancel h\cdot 4x^2(x^2+2xh+h^2)}\]

Answer 51

\[\frac{ -12x-h }{ 4x ^{2}(x ^{2}+2xh-h ^{2}) }\]?

Answer 52

That's the answer?

Answer 53

Not yet. Take the limit as h goes to zero.

Answer 54

So... that equation equals zero?

Answer 55

or do you mean take out all the h in the equation?

Answer 56

No, just let h = 0, essentially.

Answer 57

so, \[\frac{ -12x }{ 4x ^{2}(x ^{2}+2x) }\] ?

Answer 58

Redo it... there was a 2xh in your denominator :)

Answer 59

but you said h = 0?

Answer 60

Yeah, so why do you still have a 2x in your denominator? :P

Answer 61

so, 4x^2(x^2) is the answer? :p

Answer 62

for the bottom bit*

Answer 63

Tell you what? let's just abandon the entire thing of first principle and directly differenciate it.

Answer 64

If you're gonna help me with it, sure :p

Answer 65

wait, so \[\frac{ -12x }{ 4x ^{4} }\] Is the answer? :p

Answer 66

No?

Answer 67

Yes, sorry :)

Answer 68

But when you differentiate it you get a different answer?

Answer 69

I think you can simplify this
\[ \frac{ -12x }{ 4x ^{4} } \]

Answer 70

Nope. You should get \(\large -3x^{-3}\) which is equal to your answer, when simplified.

Answer 71

divide top and bottm by x and by 4

Answer 72

Okay, i think i've got it, thanks guys/girls :)