having a problem with algebraic simplification

**angypangy** · November 20th 2011, 01:06 AM

My differentiation problem is:

$y=(\frac{1}{x} - x)^3$

I get to:

$\frac{dy}{dx} = 3(x^{-1} - x)^2 . -x^{1/2}$

= $\frac{-3}{x^2} (x^{-1} - x)^2$

But book answer says:
$3(\frac{1}{x} + x)^2 (1 - \frac{1}{x^2})$

How do they get to that?

**mr fantastic** · November 20th 2011, 01:34 AM

Originally Posted by angypangy

My differentiation problem is:

$y=(\frac{1}{x} - x)^3$

I get to:

$\frac{dy}{dx} = 3(x^{-1} - x)^2 . -x^{1/2}$

= $\frac{-3}{x^2} (x^{-1} - x)^2$

But book answer says:
$3(\frac{1}{x} + x)^2 (1 - \frac{1}{x^2})$

How do they get to that?

Your calculation of dy/dx is wrong. I assume you realise that you have to use the chain rule ....?

In which case, I suggest you go back and re-do the calculation - do every step and set out your work exectly like the examples in your textbook.

If you need more help, please show ALL steps of your calculation so that specific errors, if present, can be pointed out.

**angypangy** · November 20th 2011, 02:35 AM

Originally Posted by mr fantastic

Your calculation of dy/dx is wrong. I assume you realise that you have to use the chain rule ....?

In which case, I suggest you go back and re-do the calculation - do every step and set out your work exectly like the examples in your textbook.

If you need more help, please show ALL steps of your calculation so that specific errors, if present, can be pointed out.

Ah yes I see I made a silly error. OK here is my corrected attempt.

$\frac{dy}{dx} = 3 (\frac{1}{x} - x)^2 (\frac{-1}{x^2} - 1)$

Is that correct so far?

**e^(i*pi)** · November 20th 2011, 02:40 AM

Originally Posted by angypangy

Ah yes I see I made a silly error. OK here is my corrected attempt.

$\frac{dy}{dx} = 3 (\frac{1}{x} - x)^2 (\frac{-1}{x^2} - 1)$

Is that correct so far?

Yep, that's good. It would appear the book has a sign error since they have +1 instead of -1 in the last bracket

**angypangy** · November 20th 2011, 03:07 AM

Originally Posted by e^(i*pi)

Yep, that's good. It would appear the book has a sign error since they have +1 instead of -1 in the last bracket

Ah OK, this particular book is full of errors in the answers. Thanks, so it was just a silly error in my differentiation.

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