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Differentiation

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Reply 40
Avatar for _Xenon_
_Xenon_
OP
12
Original post by notnek
Can you tell me what qualification you are doing? E.g. IGCSE Edexcel?

Then we will know if you will need to practice harder questions or not.


how can I differentiate with respect to x

1/x
Reply 41
Avatar for Zacken
Zacken
22
Original post by _Xenon_
how can I differentiate with respect to x

1/x


Rewrite it as 1x=x−1\frac{1}{x} = x^{-1} and then it's the same thing all over again.
Reply 42
Avatar for _Xenon_
_Xenon_
OP
12
Original post by Zacken
Rewrite it as 1x=x−1\frac{1}{x} = x^{-1} and then it's the same thing all over again.


-x^-2 ?
Reply 43
Avatar for Zacken
Zacken
22
Original post by _Xenon_
-x^-2 ?


Indeed. And can you write that in fraction form for the bantz?
Reply 44
Avatar for _Xenon_
_Xenon_
OP
12
Original post by Zacken
Indeed. And can you write that in fraction form for the bantz?


not sure how ?
Original post by _Xenon_
not sure how ?


x^-n = 1/(x^n)
Reply 46
Avatar for Notnek
Notnek
21
I've linked a set of IGCSE Edexcel calculus questions to give an idea of how hard the questions will be.

http://www.speedyshare.com/g9sEV/dydx.pdf

I won't be around much this afternoon so hopefully Zacken can help you with some of these if he's free.

Zacken, I think you did CIE IGCSE? So I don't know if you are aware of the IGCSE spec but these questions should help.
Reply 47
Avatar for Zacken
Zacken
22
Original post by notnek
I've linked a set of IGCSE Edexcel calculus questions to give an idea of how hard the questions will be.

http://www.speedyshare.com/g9sEV/dydx.pdf

I won't be around much this afternoon so hopefully Zacken can help you with some of these if he's free.

Zacken, I think you did CIE IGCSE? So I don't know if you are aware of the IGCSE spec but these questions should help.


Cheers, thanks for the link, just downloaded it. I'll be around for most of the evening, so I should be able to help. @_Xenon_ - mind downloading the pdf he's linked and then we can go through the questions together? Post a picture of your answers or ask for help with any of them (I've got a copy, so you just need to reference the question number) and I can verify your answers and/or help out with troubles you have.
Reply 48
Avatar for _Xenon_
_Xenon_
OP
12
Original post by Zacken
Cheers, thanks for the link, just downloaded it. I'll be around for most of the evening, so I should be able to help. @_Xenon_ - mind downloading the pdf he's linked and then we can go through the questions together? Post a picture of your answers or ask for help with any of them (I've got a copy, so you just need to reference the question number) and I can verify your answers and/or help out with troubles you have.


OK thanks man. :smile:
Reply 49
Avatar for Zacken
Zacken
22
Original post by _Xenon_
OK thanks man. :smile:


No worries, when do you want to start? :smile:
Reply 50
Avatar for _Xenon_
_Xenon_
OP
12
Original post by Zacken
No worries, when do you want to start? :smile:


just downloaded and now I'll try them, thanks.
Reply 51
Avatar for Zacken
Zacken
22
Original post by _Xenon_
just downloaded and now I'll try them, thanks.


Awesome - just post your answers here and I'll verify them whenever I can!
Reply 52
Avatar for _Xenon_
_Xenon_
OP
12
Original post by Zacken
Awesome - just post your answers here and I'll verify them whenever I can!


First page:

http://i.imgur.com/b7gH0HX.jpg
Original post by _Xenon_


Firstly, notation wise, on the last question y is not =3X^2 - Y=x^3. You want to write is as dy/dx =3x^2, or you can also write y'=3x^2. Once you have the expression for the gradient, you should be able to work out the values of x that satisfy the conditions, and then you can use the original equation (y=x^3 to get the coordinates)
Reply 54
Avatar for Zacken
Zacken
22
Original post by _Xenon_


First two is correct, well done!

The gradient at any point (a,b)(a,b) on a curve y=f(x)y = f(x) is given by finding dydx∣x=a\frac{\mathrm{d}y}{\mathrm{d}x} \bigg|_{x=a}.

So, for example, the gradient at (3,9)(3,9) on the curve y=x2y = x^2 is given by me differentiating the equation of the curve to get 2x2x and then plugging x=3x=3 into this derivative to get gradient at (3,9) = 2×3=62 \times 3 = 6.

Now, if I want to find the point at which the gradient of the curve is a certain value, I just take my derivative and set it equal to that value and solve the equation. So, for example, if I want the point at which the gradient of y=x2y=x^2 is 1212 then I take the derivative 2x2x (this is my gradient at any point x), but I want the gradient to be 12, so 2x=12⇒x=62x = 12 \Rightarrow x = 6.

This is the x-coordinate of the point at which the gradient is 12; the y-coordinate is given by plugging my x into my function to get y = 6^2 = 36. So the point (6, 36) on the curve y = x^2 has gradient 12.

What is the gradient at (2, 8) of the curve y=x3y = x^3?

What point of the curve y=x3y=x^3 has gradient 2727?

Once you've done this for me, can you now do the question in the booklet?

But yeah, when you find the derivative, you should not be doing:

y = x^3
y = 3x^2

You should do:

y=x^3
dy/dx = 3x^2

They mean very different things and you should be careful as to how you write them.
(edited 7 years ago)
Reply 55
Avatar for _Xenon_
_Xenon_
OP
12
Original post by samb1234
Firstly, notation wise, on the last question y is not =3X^2 - Y=x^3. You want to write is as dy/dx =3x^2, or you can also write y'=3x^2. Once you have the expression for the gradient, you should be able to work out the values of x that satisfy the conditions, and then you can use the original equation (y=x^3 to get the coordinates)


Oops that was what I meant, sorry. I was supposed to write dy/dx but wrote y.

So 2^3 = 8? Then what
Reply 56
Avatar for Zacken
Zacken
22
Original post by _Xenon_
Oops that was what I meant, sorry. I was supposed to write dy/dx but wrote y.

So 2^3 = 8? Then what


So one point is (2, 8)

Now, do you remember how to solve equations of the form x2=4x^2 = 4? One solution is x=4=2x = \sqrt{4} = 2. What about the other solution?

Edit: What two numbers can you square to give you 4?
(edited 7 years ago)
Original post by _Xenon_
Oops that was what I meant, sorry. I was supposed to write dy/dx but wrote y.

So 2^3 = 8? Then what


You're missing one value of x...
Reply 58
Avatar for _Xenon_
_Xenon_
OP
12
Original post by Zacken
So one point is (2, 8)

Now, do you remember how to solve equations of the form x2=4x^2 = 4? One solution is x=4=2x = \sqrt{4} = 2. What about the other solution?

Edit: What two numbers can you square to give you 4?


Is the other one negative? So (-2,-8)
Reply 59
Avatar for Zacken
Zacken
22
Original post by _Xenon_
Is the other one negative? So (-2,-8)


Yep! Well done. :biggrin:

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