Beginner calculus question... (1 Viewer)

[OH1995]

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????

 

[powlmao]

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

x=1

Therefore

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

 

[Siddy123]

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

 

[powlmao]

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

 

[powlmao]

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

 

[powlmao]

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

 

[Siddy123]

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...

 

[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

 

[xdosha]

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

 

[powlmao]

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


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

 

[Siddy123]

Yes, sorry saw it wrong, wow I couldn't even do a basic 2u question.
oh lord, my hsc LOL

dw cuz.
YOLO.

 

[powlmao]

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

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

 

[xdosha]

dw cuz.
YOLO.

YES YOLO!
are you doing 4u and 3u hsc this year? and you're in year 11?

 

[Siddy123]

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?

 

[Sy123]

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

 

[powlmao]

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

 

[xdosha]

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!

 

[Siddy123]

nice name m8
and u gotta teach me how to use LaTex one time

 

[powlmao]

FYI

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

 

[Siddy123]

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