Solve Calculus Derivative: Step-by-Step Guide for f'(3x-6)/x

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Hi,

I am just starting out in calculus, and I'm not sure how to work through this type of question:

f' \frac{3x-6}{x}

So I have:

(3x-6)(x^{-1})
then (3x -6)(-x^{-2})

Now what?

(3)(-x^{-2})?

The book gives an answer of 6(x^{-2})

Thanks in advance
 
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The first thing id probably do is break down that fraction into 3-\frac{6}{x} = 3 - 6x^{-1} and then work from there.
 
Aha, got it, thank you!
 
Cmunro said:
Hi,

I am just starting out in calculus, and I'm not sure how to work through this type of question:

f' \frac{3x-6}{x}

So I have:

(3x-6)(x^{-1})
then (3x -6)(-x^{-2})

I'm guessing you haven't gotten to the Quotient Rule yet. When you have, you'll see that you could also work this out using that (and you'll get the same answer), but you won't get the result you have on your last line here...

You could also use the Product Rule on your expression
(3x-6)(x^{-1}), which would give you two terms, the one you found:

(3x -6)(-x^{-2})

plus a second term

(3x-6)' · (x^{-1}) = 3 · (x^{-1}) ,

adding up to (3x -6-3x)(-x^{-2}) = 6/(x^{-2}) .

All three methods, the one you ultimately used and these two, give the same answer.
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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