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+33666

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.

AlanApr 3, 2014

+33666

Rom and Reinout have answered the first part, so for the second part:

dydx2.PNG

AlanApr 3, 2014

+2354

Melody:
140403 garage door derivative.JPG

I see ROM already answered in the meanwhile,

for the second part, it could also be that d&/dx = 0

Reinout

reinout-gApr 3, 2014

+2354

Melody:
140403 garage door derivative.JPG

Let me define theta with this symbol &

I think what is the case here is that & is a fuction of x.

Then lets split y = xcot& + 1 - 2fcos&
into
y = y ₁ + y ₂ where y ₁ = xcot& and y ₂ = 1 - 2fcos&

then dy ₁/dx = cot& + x*d&/dx(cot&) = cot& + x* -csc ²& * d&/dx

Note that: -csc ²x = -(cscx) ² = -(1/(sinx)) ² = -1/(sin ²x), because cscx = 1/(sinx)

Therefore we have;

dy ₁/dx = cot & + x * -1/(sin ²x) * d&/dx

then for y ₂ we have

dy ₂/dx = 2fsin& * d&/dx

then dy/dx = 2fsin& * d&/dx + cot & + x * -1/(sin ²x) * d&/dx = cot & + (2fsin& - x/(sin ²x))d&/dx

Let me post this first before I look at the second part of your question

reinout-gApr 3, 2014

+6252

Clipboard01.jpg

RomApr 3, 2014

+118725

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.

MelodyApr 3, 2014

+118725

140403 factor tree.JPG

MelodyApr 3, 2014

+118725

help me:
(c^5a/c^4a)^1

This question is open to interpretation

[size=150]It is really
[ (c ⁵*a/c ⁴) * a ] ¹= ca ²
[/size]
But I think this may be what you meant.

140403 indices.JPG

MelodyApr 3, 2014

+2354

Stu:
Hi, test question here, (with numbers). I'm after the basic concept and methods of nx³-nx²+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 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

reinout-gApr 3, 2014

+118725

Thank you Alan

MelodyApr 3, 2014

+33666

Stu:
Hi, test question here, (with numbers). I'm after the basic concept and methods of nx³-nx²+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 ²-2nx1) The equation of the tangent line is then y - y1 = m(x - x1).

AlanApr 3, 2014

+1313

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.

StuApr 3, 2014

+1313

Hey, start using the forum in the way that is respectful to the people who answer your questions @op. Google what OP means.

StuApr 3, 2014

+33666

esketc:
If $2000 is invested at an interest rate of 6.5% per year, compounded daily, find the value of the investment after the given number of years. (Round your answers to the nearest cent.)

The future value, FV, is given by FV = P(1 + r/n) ^n*t where P is the principal investment ($2000) r is the annual interest rate, expressed as a decimal fraction rather than a percentage (0.065), n is the number of times it's compounded per year (365) and t is the number of years.

So

FV=2000*(1+0.065/365)^(365*t)

You don't say what t is, so, if we assume t is 5 years, then the future value is

2000*(1+0.065/365)^(365*5)

This is $2767.98 to the nearest cent.

AlanApr 3, 2014

+33666