# Thread: two Q in derivative Calculation by using prouduct rule

1. ## two Q in derivative Calculation by using prouduct rule

Hi all

Two Q in derivative Calculation by using prouduct rule

see the attachment

2. EDIT: To clarify, I am answering part 2.

When dealing with derivatives, the most important thing is to be careful. There are dropped and confused signs, arithmatic errors, and your substitution for v is wrong. (Also, not that it matters, but you kind of pulled your last step out of nowhere: continuing from the step above would give 2x). I'm sorry to say but I have to scrap everything except the mini-derivatives; those were correct.

u'v + v'u =

$(1-\frac{1}{x^2})(x-\frac{1}{x}+1) \, \, + \, \, (1+\frac{1}{x^2})(x+\frac{1}{x})$

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

$x-\frac{2}{x}+1+\frac{1}{x^3}-\frac{1}{x^2}+x + \frac{2}{x} + \frac{1}{x^3}$

$2x+1-\frac{1}{x^2}+\frac{2}{x^3}$

3. Q1 should be $+12x^2$ instead of $+6x^2$

Second last line of that answer you had $-2x^2$ and $-x^2$ instead of +

4. Originally Posted by r-soy
Hi all

Two Q in derivative Calculation by using prouduct rule

see the attachment
1. You got the sign of x² wrong. See attachment.

2. $y=\left(x+\frac1x \right)\left(x-\frac1x +1 \right)$

Then

$y'=\left(1-\frac1{x^2} \right)\left(x-\frac1x +1 \right) + \left(x+\frac1x \right)\left(1+\frac1{x^2} \right)$

Expand the brackets carefully, collect like terms. The final result contains 4 summands.

5. Originally Posted by Turiski
EDIT: To clarify, I am answering part 2.

When dealing with derivatives, the most important thing is to be careful. There are dropped and confused signs, arithmatic errors, and your substitution for v is wrong. (Also, not that it matters, but you kind of pulled your last step out of nowhere: continuing from the step above would give 2x). I'm sorry to say but I have to scrap everything except the mini-derivatives; those were correct.

u'v + v'u =

$(1-\frac{1}{x^2})(x-\frac{1}{x}+1) \, \, + \, \, (1+\frac{1}{x^2})(x+\frac{1}{x})$

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

$x-\frac{2}{x}+1+\frac{1}{x^3}-\frac{1}{x^2}+x + \frac{2}{x} + \frac{1}{x^3}$

$2x+1-\frac{1}{x^2}+\frac{2}{x^3}$

in multimpling why you said (-1/x^2 ) (x) =
-1/x not -1/x^-1

???

6. Originally Posted by r-soy
in multimpling why you said (-1/x^2 ) (x) =
-1/x not -1/x^-1

???
$\frac{-1}{x^2}\cdot x = \frac{-x}{x^2} = \frac{-1}{x} \neq \frac{-1}{x^{-1}} = -x$