1. Beginner calculus question...

For y=x3 find dy/dx at x=1

I did this. How do I find the answer, because i probably did it wrong.

dy/dx= lim 1-->c (13-c3)/1-c
=lim 1-->c (1-c)(c2+c+1)/1-c
=lim 1-->c c2+c+1
how do i get c????

2. Re: Beginner calculus question...

y= x^3
y'= 3x^2

x=1

Therefore

F'(x) = 3x^2
f'(1)= 3(1)^2
f'(1)= 3

3. Re: Beginner calculus question...

u can differentiate it normally if u want( simpler) but it is good to know first principles.
so y=x^3 ( when u differentiate u bring the power down and times it by the coefficient and u minus one from original power..)
so dy/dx=3x^2
at x=1
sub it in.
,=3(1)^2
=3

4. Re: Beginner calculus question...

Where did you get c from? It looks like you are using first principal. Use short method it is easier.

This should take max three steps.

State it
Differeniate it
Sub in x and solve

5. Re: Beginner calculus question...

Originally Posted by xdosha
if you sub 1 into f'
shouldn't it be, 3(1)^2
which is 9?
No, it is 3

1 square times 3 is 3

6. Re: Beginner calculus question...

Originally Posted by xdosha
if you sub 1 into f'
shouldn't it be, 3(1)^2
which is 9?

You're doing it in the wrong order.

You solve powers first then mulitply

7. Re: Beginner calculus question...

Originally Posted by xdosha
if you sub 1 into f'
shouldn't it be, 3(1)^2
which is 9?
the squared is only on the xvalue... and then u times by 3
but if it was (3x)^2
then u would be right...

8. Re: Beginner calculus question...

Yeh we just started today and have learnt the first principle. How do I do it using the long way, because we haven't done differentiation properly yet :P

9. Re: Beginner calculus question...

Originally Posted by Siddy123
the squared is only on the xvalue... and then u times by 3
but if it was (3x)^2
then u would be right...
Yes, sorry saw it wrong, wow I couldn't even do a basic 2u question.
oh lord, my hsc LOL

10. Re: Beginner calculus question...

Originally Posted by xdosha
Yes, sorry saw it wrong, wow I couldn't even do a basic 2u question.
oh lord, my hsc LOL
lol it would be a year 8 question haha and you are doing ext 2 :P

Its alright. Its not a bad mistake, just one of those things people over look.

11. Re: Beginner calculus question...

Originally Posted by xdosha
Yes, sorry saw it wrong, wow I couldn't even do a basic 2u question.
oh lord, my hsc LOL
dw cuz.
YOLO.

12. Re: Beginner calculus question...

Originally Posted by OH1995
Yeh we just started today and have learnt the first principle. How do I do it using the long way, because we haven't done differentiation properly yet :P
Long way is first principal

The short method is for only equations like 3x^2 (ones which aren't fractions, brackets etc.)

You take the power put it at the front and then reduce the power again.

i.e. x^3

Take the power of three to the front and reduce the power to 2 to make 3x^2

13. Re: Beginner calculus question...

Originally Posted by Siddy123
dw cuz.
YOLO.
YES YOLO!
are you doing 4u and 3u hsc this year? and you're in year 11?

14. Re: Beginner calculus question...

Originally Posted by xdosha
YES YOLO!
are you doing 4u and 3u hsc this year? and you're in year 11?
damn son, u do 4unit and u can read a sig.
marry me pls.
hehehe, nah, on a srs note.
yeah i do u do 4u as well aye how u finding it?

15. Re: Beginner calculus question...

$f(x)=x^3 \\ f(x+h)=(x+h)^3 \\ \\ \frac{dy}{dx}=\lim_{h\rightarrow 0} \frac{(x+h)^3-x^3}{h} \\ \\ = \lim_{h \rightarrow 0} \frac{3x^2h+3xh^2+h^3}{h} \\ \\ =\lim_{h \rightarrow 0} 3x^2+3xh+h^2 \\ \\ =3x^2 \\ \\ \therefore \frac{dy}{dx}=3x^2 \\ \\ x=1 \Rightarrow \frac{dy}{dx} = 3(1)^2=3$

I hope you know, that I just did not bother writing down the expansion then simplifying etc.

16. Re: Beginner calculus question...

I hate using first principal. It is such an effort .

17. Re: Beginner calculus question...

Originally Posted by Siddy123
damn son, u do 4unit and u can read a sig.
marry me pls.
hehehe, nah, on a srs note.
yeah i do u do 4u as well aye how u finding it?
pretty hard, but it's okay I guess?
good luck in the HSC, i'm competing against you, go ez!

18. Re: Beginner calculus question...

Originally Posted by Sy123
$f(x)=x^3 \\ f(x+h)=(x+h)^3 \\ \\ \frac{dy}{dx}=\lim_{h\rightarrow 0} \frac{(x+h)^3-x^3}{h} \\ \\ = \lim_{h \rightarrow 0} \frac{3x^2h+3xh^2+h^3}{h} \\ \\ =\lim_{h \rightarrow 0} 3x^2+3xh+h^2 \\ \\ =3x^2 \\ \\ \therefore \frac{dy}{dx}=3x^2 \\ \\ x=1 \Rightarrow \frac{dy}{dx} = 3(1)^2=3$
nice name m8
and u gotta teach me how to use LaTex one time:P

19. Re: Beginner calculus question...

FYI

For you 4U students, i uploads a heap of resources for you in the 4u thread

20. Re: Beginner calculus question...

Originally Posted by xdosha
pretty hard, but it's okay I guess?
good luck in the HSC, i'm competing against you, go ez!
thank you
all the best to you.
and yeah, me and only the rest of NSW.
ill make sure i let every1 know to take it ezy aye

21. Re: Beginner calculus question...

Originally Posted by Siddy123
nice name m8
and u gotta teach me how to use LaTex one time:P
Haha, yes I like your name too.

I gradually learnt latex through: http://www.codecogs.com/latex/eqneditor.php

I kept using the buttons to make the image, and gradually I learnt all the commands by heart, and now I just use the tex /tex editor here

22. Re: Beginner calculus question...

Originally Posted by Sy123
Haha, yes I like your name too.

I gradually learnt latex through: http://www.codecogs.com/latex/eqneditor.php

I kept using the buttons to make the image, and gradually I learnt all the commands by heart, and now I just use the $ " /> built in editor here.
There's an in built editor here? I dont see the 'function' button that they use to have....

23. Re: Beginner calculus question...

Originally Posted by Sy123
Haha, yes I like your name too.

I gradually learnt latex through: [URL]http://www.codecogs.com/latex/eqneditor.php[/URL]

I kept using the buttons to make the image, and gradually I learnt all the commands by heart, and now I just use the tex /tex editor here
ill check that out.
thanx

24. Re: Beginner calculus question...

Originally Posted by Peeik
There's an in built editor here? I dont see the 'function' button that they use to have....

Well you have to bound the commands with the commands

[$tex] and at the end [/tex$]

There are NO dollar signs here. I just did that otherwise it will think Im writing latex

25. Re: Beginner calculus question...

Originally Posted by Sy123
Well you have to bound the commands with the commands

[$tex] and at the end [/tex$]

There are NO dollar signs here. I just did that otherwise it will think Im writing latex
ahhhhh i see what you mean now.....and you just put the latex coding in between there?

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