Hi, test question here, (with numbers). I'm after the basic concept and methods of nx ^{3}-nx ^{2}+c, I figure that y=mx+c is required but not sure how to get that even when I plug in the value for x1 for example.

Stu Apr 3, 2014

#1**0 **

Actually there is a lot of help i need with derivatives and this topic, like tangents, intervals and all sorts of things in this test. Any suggestion, when I read the info in 2 hours i gain no understanding and no closer to solving it.

Stu Apr 3, 2014

#2**0 **

I'm not entirely sure what you are asking, but if you mean how do you find the equation of the tangent line to your cubic at some point (x1, y1) on the cubic, then the gradient of the tangent will just be given by differentiating the cubic and plugging x1 in. (m = 3nx1^{2}-2nx1) The equation of the tangent line is then y - y1 = m(x - x1).

Stu:Hi, test question here, (with numbers). I'm after the basic concept and methods of nx

^{3}-nx^{2}+c, I figure that y=mx+c is required but not sure how to get that even when I plug in the value for x1 for example.

I'm not entirely sure what you are asking, but if you mean how do you find the equation of the tangent line to your cubic at some point (x1, y1) on the cubic, then the gradient of the tangent will just be given by differentiating the cubic and plugging x1 in. (m = 3nx1

Alan Apr 3, 2014

#3**0 **

Hey Stu,

so if I'm correct, what you're trying to find is the tangent of a function of the following form; nx^{3}-nx ^{2}+c?

First find the derivative of nx^{3}-nx ^{2}+c

df/dx = 3nx^{2}-2nx.

Now the tangent at f(x*,y*) is given by y = (df/dx*)x + c, where (x*,y*) is the point where you need to calculate the tangent

Then, to find c, we solve y* = (df/dx*)x* + c.

To give an idea look at the following example.

suppose we want to calculate the tangent of f(x) for x=1

then y = 4*1^{3}-2*1 ^{2}+6 = 8

df/dx = 12x^{2}-4x

so df/dx* = 12*1^{2}-4*1 = 8

the tangent must then be given by y = 8x + c

fill in the point (x*,y*) = (1, 8) => 8 = 8*1 + c => c = 0

hence the tangent at (1,8) is given by y = 8x

Stu:Hi, test question here, (with numbers). I'm after the basic concept and methods of nx

^{3}-nx^{2}+c, I figure that y=mx+c is required but not sure how to get that even when I plug in the value for x1 for example.

Hey Stu,

so if I'm correct, what you're trying to find is the tangent of a function of the following form; nx

First find the derivative of nx

df/dx = 3nx

Now the tangent at f(x*,y*) is given by y = (df/dx*)x + c, where (x*,y*) is the point where you need to calculate the tangent

Then, to find c, we solve y* = (df/dx*)x* + c.

To give an idea look at the following example.

suppose we want to calculate the tangent of f(x) for x=1

then y = 4*1

df/dx = 12x

so df/dx* = 12*1

the tangent must then be given by y = 8x + c

fill in the point (x*,y*) = (1, 8) => 8 = 8*1 + c => c = 0

hence the tangent at (1,8) is given by y = 8x

reinout-g Apr 3, 2014

#4**0 **

Hi Stu,

Ok I'll do some examples maybe that will help

Say you want to differentiate

1) y = x^{7}

dy/dx = 7x^{7-1} = 7x ^{6}

This is a formula for the gradient of the tangent.

If you want to know what the gradient of the tangent is when x=2 you would substitute in x=2 in and get dy/dx = 7 * 2^{6} = 448 (at x=2)

2) y = 9x^{12}

You bring the power down the front and multiply it by any coefficient that is already there -> put the x -> then drop the power down by one.

dy/dx = 12*9x^{11}

dy/dx = 108x^{11}

If you want to know the gradient of the tangent at the point (1,9) then you will have to substitute x=1 into the derivative formula. dy/dx = 108 [at (1,9)}

It is always the x value that you sub in.

I'll show you how you can use different notation. I know that they have been confusing you.

3) d/dx 5x^{4} = 20x ^{3}

4) f(x) = -4x^{3}

f '(x) = -12x^{2}

5) y = 7x^{-2}

y' = -14x^{-3}

In this case y' is a short cut notation for dy/dx

NOW If there are a lot of terms ADDED or SUBTRACTED together you can just do each one individually.

6)y = 3x^{5} - 4x ^{2} + 9x + 23

dy/dx = 5*3x^{5-1} - 2*4x ^{2-1} + 9 + 0

dydx = 15x^{4} - 8x ^{1} + 9

dy/dx = 15x^{4} - 8x + 9

7) y=mx+c

dy/dx = m because x has a power of 1 and 1-1=0 and x^{0}=1

also remember that I told you the derivative gives you the gradient of the tangent to a curve. A straight line always has the same gradient. It is m.

8) d/dx ( nx^{3}-nx ^{2}+c ) = 3nx ^{2} - (n-1)x ^{1} + 0 => 3nx ^{2} - (n-1)x

I hope that this helps.

Ok I'll do some examples maybe that will help

Say you want to differentiate

1) y = x

dy/dx = 7x

This is a formula for the gradient of the tangent.

If you want to know what the gradient of the tangent is when x=2 you would substitute in x=2 in and get dy/dx = 7 * 2

2) y = 9x

You bring the power down the front and multiply it by any coefficient that is already there -> put the x -> then drop the power down by one.

dy/dx = 12*9x

dy/dx = 108x

If you want to know the gradient of the tangent at the point (1,9) then you will have to substitute x=1 into the derivative formula. dy/dx = 108 [at (1,9)}

It is always the x value that you sub in.

I'll show you how you can use different notation. I know that they have been confusing you.

3) d/dx 5x

4) f(x) = -4x

f '(x) = -12x

5) y = 7x

y' = -14x

In this case y' is a short cut notation for dy/dx

NOW If there are a lot of terms ADDED or SUBTRACTED together you can just do each one individually.

6)y = 3x

dy/dx = 5*3x

dydx = 15x

dy/dx = 15x

7) y=mx+c

dy/dx = m because x has a power of 1 and 1-1=0 and x

also remember that I told you the derivative gives you the gradient of the tangent to a curve. A straight line always has the same gradient. It is m.

8) d/dx ( nx

I hope that this helps.

Melody Apr 3, 2014

#5**0 **

There is a lot of useful material at http://www.mathcentre.ac.uk/ that might be of some help to you.

Stu:Actually there is a lot of help i need with derivatives and this topic, like tangents, intervals and all sorts of things in this test. Any suggestion, when I read the info in 2 hours i gain no understanding and no closer to solving it.

There is a lot of useful material at http://www.mathcentre.ac.uk/ that might be of some help to you.

Alan Apr 3, 2014

#7**0 **

I think in part the online test allows only for the correct answer and I do not have enough knowledge to put it in correctly, in answering the question. That if I try I am not told correct parts or inccorect parts and I am given knew equations if I fail. Thus I must investigate the right answers since I could be either a little wrong or majorly wrong etc. Thanks.

Stu Apr 4, 2014

#8**0 **

You mean during practice tests? Can you save your calculations somewhere or write them down? Then you can check what you did at a later moment or post some of the things you can't work out on the forum. That could help you progress.

Stu:I think in part the online test allows only for the correct answer and I do not have enough knowledge to put it in correctly, in answering the question. That if I try I am not told correct parts or inccorect parts and I am given knew equations if I fail. Thus I must investigate the right answers since I could be either a little wrong or majorly wrong etc. Thanks.

You mean during practice tests? Can you save your calculations somewhere or write them down? Then you can check what you did at a later moment or post some of the things you can't work out on the forum. That could help you progress.

reinout-g Apr 4, 2014