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# AS-Level Maths - Differentiation help please! watch

1. Hiya,

I've just got down to starting some homework from maths and don't really understand how to do some of the questions for it. I've already got the answers in the back of the book, but I don't really know how they got to it and wanted to know if anyone could explain.

1) Rewrite these equations using index notation, and then differentiate Y with respect to X.

g) y= -(1/x^2)

The answer in the book is 2/x^3 but i don't really know how they got to it. If anyone could help, thanks!

The other question is this one:

5) Given y=f(x) find f'(x) in each case:
g) f(x) = ((x+3)^2) / x

The answer is apparently 1 - (9/x^2) but again I don't really know how they got to it. If you could explain the step-by-step process of it at all, I would really appreciate it.

2. 1.
g) rewritten in index notation is -1x^-2.
This is then differentiated to 2x^-3 which is equivalent to 2/x^3

5.
g) expand brackets to get (x^2+6x+9)/x
Which then becomes x+6+9x^-1 due to rules of indicies.
Which is then differentiated to 1-9x^-2 or 1-(9/x^2) as your book says.

Hope this helps.

If you need more explanation I can help just didn't want to go into too much detail incase you understand it all already and just needed a pointer.
Source: did AS last year and got an A
3. (Original post by Dubois98)
Hiya,

I've just got down to starting some homework from maths and don't really understand how to do some of the questions for it. I've already got the answers in the back of the book, but I don't really know how they got to it and wanted to know if anyone could explain.

1) Rewrite these equations using index notation, and then differentiate Y with respect to X.

g) y= -(1/x^2)

The answer in the book is 2/x^3 but i don't really know how they got to it. If anyone could help, thanks!

The other question is this one:

5) Given y=f(x) find f'(x) in each case:
g) f(x) = ((x+3)^2) / x

The answer is apparently 1 - (9/x^2) but again I don't really know how they got to it. If you could explain the step-by-step process of it at all, I would really appreciate it.

Has your teacher thoroughly explained the concept of dy/dx (or "f'(x)")?

If you can write y as a series of powers of x - for example,

y = (ax^4) + (bx^3) + (cx^2) +(dx) + (e) + (fx^-1) + (gx^-2) + (hx^-3),

then (noting that e = ex^0), the differential will be

(4ax^3) + (3bx^2) + (2cx) + (1d) + (0) + (-1fx^-2) + (-2gx^-3) + (-3hx^-4).

The manipulation is therefore straightforward provided you can write the equation for y in the appropriate format.
4. (Original post by Bjc00)
1.
g) rewritten in index notation is -1x^-2.
This is then differentiated to 2x^-3 which is equivalent to 2/x^3

5.
g) expand brackets to get (x^2+6x+9)/x
Which then becomes x+6+9x^-1 due to rules of indicies.
Which is then differentiated to 1-9x^-2 or 1-(9/x^2) as your book says.

Hope this helps.

If you need more explanation I can help just didn't want to go into too much detail incase you understand it all already and just needed a pointer.
Source: did AS last year and got an A

Thanks for the response - I understand the first question, but am still a bit stumped on the 2nd. Why does it become x+6+9x^-1? Should it not be something more like 2x+6+0? or am I completely at the wrong end of the stick?
5. (Original post by Dubois98)
Thanks for the response - I understand the first question, but am still a bit stumped on the 2nd. Why does it become x+6+9x^-1? Should it not be something more like 2x+6+0? or am I completely at the wrong end of the stick?
To make it into an easier form to deal with get it all on 1 line. When x^2+6x+9 is all divided by x the powers are taken away from each other. x is x^1 and so x^2 becomes x^1 (or just x), 6x becomes 6x^0 (or just 6) and the 9 becomes 9x^-1. then differentiate x+6+9x-1.
Hope that's easier to see now.

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