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**ninjaman****Member**- Registered: 2013-10-15
- Posts: 53

hello

I have a question im stuck with

y = (7x-4) 3/2 (3/2 is supposed to be 3 over 2, I don't know how to do that on here)

I got this,

u = 7x-4

du/dx = 7

y = u 3/2

dy/du = 3/2u -3/2 (the -3/2 is power, supposed to be above right of u)

du/dx * dy/du = 3/2u-3/2 * 7 = 21/2 u -3/2

since u = 7x-4, 21/2 (7x-4) -3/2

im thinking most of it is correct except for the -3/2, the fraction part has me concerned.

thanks

simon:)

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 103,772

Hi;

ninjaman wrote:

dy/du = 3/2u -3/2 (the -3/2 is power, supposed to be above right of u)

You made a mistake in the above step. You did not apply the power rule correctly. Here are the steps for the whole problem.

You can use latex for all your math posting. I use this when I am not doing it by hand,

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.** **A number by itself is useful, but it is far more useful to know how accurate or certain that number is.**

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**ninjaman****Member**- Registered: 2013-10-15
- Posts: 53

thanks a lot, I looked at the bit where you put in a radical(I think that's what its called, the sqrt sign) that threw me.

thanks also for the latex thing, that will help a lot when writing out maths stuff!

all the best

simon

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 103,772

Hi;

It makes perfect latex so everything looks nice and neat.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.** **A number by itself is useful, but it is far more useful to know how accurate or certain that number is.**

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