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Ricky
2006-01-10 02:28:24

Just to clarify:

Now we have:

This isn't exactly clear, so let's write it in a different way:

Or:

Or even:

Is that clearer?

RickyOswaldIOW
2006-01-09 18:22:31

Why does 1/(4x) = 4^-1 * x^-1?   I was taught to recirpocal i.e. simply bring the bottom of the fraction to the top and change the sign of the power.  Why have you split it into 4^-1 * x^-1?

P.S. thanks for explaining the fractions, it makes a lot of sense.

Ricky
2006-01-09 16:48:49

(a / b) / (c / d) = (a / b) * (d / c)

(3 / 2) / (4 / 5) = (3 / 2) * (5 / 4) = 15 / 8

Be careful writing 1/4x.  It looks like (1/4)x, when in reality it's 1 / (4x).

y = 4 + 1/4x = 4 + 4^-1 * x^-1, that 4 is on the bottom as well.  Of course, 4^-1 = 1/4

y = 4 + 1/4 * x^-1

Try doing that.

RickyOswaldIOW
2006-01-09 15:41:00

in addition, could you check this (I think the book has put the answers in the wrong order)

y = 4 + 1/4x  at  x = -1
= 4 + 4x^-1

dy/dx = -4x^-2 = -4/x^2 = -4/1 = -4.   i.e. the Gradient of the Tangent on the curve at the stated point (x = -1) is -4 and thus the Gradient of the Normal is 1/4.  My book claims it is the other way round and with opposite signs (tang is -1/4 and norm is 4)!!

RickyOswaldIOW
2006-01-09 15:14:51

Could you explain how to work out a fraction over a fraction?

#### Ricky wrote:

-1 / (1 / 4) = -4

I've never actually been taught this

Ricky
2006-01-07 15:37:53

Nope, you're just plugging in wrong.

-x^-2 = -1 / x^2

So if x = 1/2:

-1 / (1/2)^2 = -1 / (1 / 4) = -4

RickyOswaldIOW
2006-01-07 09:10:22

I'm a little stuck again:
find the gradiaent of the tangent and the normal to the following curves at the stated point.

y = x + 1/x  at x = 1/2.

dy/dx = -x^-2 + 1  = 3/4,  thus the gradient of the normal is -4/3.

I can work out most of these sums, I am doing somthing wrong with the negative power...

krassi_holmz
2006-01-05 20:45:33

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RickyOswaldIOW
2006-01-05 15:20:49

Thanks again, one final question, the only one left that I am unsure of my workings:

y = sqrt(4/x^3) = sqrt(4x^-3) = 2x^-3/2
dy/dx = -3x^-5/2
Is this right?

Ricky
2006-01-05 15:12:39

Yes, but it is normally stated in a different way, and thus involving less theorms.

y = a * f(x), y' = a * f'(x)

So:

y = 3 * (x^4), where f(x)=x^4
y' = 3 * (4x^3) = 12x^3

RickyOswaldIOW
2006-01-05 14:34:37

in a question like y = 3x^4,  do you multiply the 4 by the 3 to make dy/dx = 12x^3?
i.e.  y=ax^b   dy/dx = (ba)x^b-1

RickyOswaldIOW
2006-01-05 14:30:40

I was subtracting 1... but from +4 and not -4
Thanks.

Ricky
2006-01-05 14:29:09

On the last one, you need to subtract, not add, 1, so you get:

4x^-5

Otherwise, they're all right.

RickyOswaldIOW
2006-01-05 13:42:19

Hi guys and girls, long time no see.  I've been away for the Christmas holidays (away from my computer at least) but now I am back to bug you with more laughably simple questions!

I've moved on to the topic of calculus and am having a little trouble with some differentiation questions.  I understand how to do basic questions such as y=x^10   dy/dx = 10x^9.
I think these are also correct, only my book shows the answers in a different format, I'd be obliged if you could check.

y = 1/x = x^-1    dy/dx = -x^-2    (book says: -1/x^2)

y=sqrt x = x^1/2    dy/dx = 1/2x^-1/2    (book says: 1/ 2sqrt x)

y= -1/x^4 = -x^-4  dy/dx = 4x^-3      (book says: 4/x^5)

P.S. Happy New Year!