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**RickyOswaldIOW****Member**- Registered: 2005-11-18
- Posts: 212

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!

Aloha Nui means Goodbye.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

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

4x^-5

Otherwise, they're all right.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**RickyOswaldIOW****Member**- Registered: 2005-11-18
- Posts: 212

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

Thanks.

Aloha Nui means Goodbye.

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**RickyOswaldIOW****Member**- Registered: 2005-11-18
- Posts: 212

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

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

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

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**RickyOswaldIOW****Member**- Registered: 2005-11-18
- Posts: 212

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?

*Last edited by rickyoswaldiow (2006-01-04 17:10:45)*

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,906

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IPBLE: Increasing Performance By Lowering Expectations.

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**RickyOswaldIOW****Member**- Registered: 2005-11-18
- Posts: 212

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...

Aloha Nui means Goodbye.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

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

*Last edited by Ricky (2006-01-06 16:38:26)*

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**RickyOswaldIOW****Member**- Registered: 2005-11-18
- Posts: 212

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

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**RickyOswaldIOW****Member**- Registered: 2005-11-18
- Posts: 212

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)!!

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

(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.

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**RickyOswaldIOW****Member**- Registered: 2005-11-18
- Posts: 212

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.

*Last edited by rickyoswaldiow (2006-01-08 19:25:40)*

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

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?

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