# Math Help - Trouble Finding Derivative

1. ## Trouble Finding Derivative

I've been working on optimization problems and have been having problems with this one. The only thing that I'm confused about is what's going on when I find the derivative of the function:

$\frac{d}{dx}[(x+d)^2(\frac{w^2}{x^2} +1)]$

Also, in case it matters, $d=24$ and $w=6$, but in the example, the derivative is found with respect to the variables rather than their respective values.

My attempt:

$[(2)(x+d) \cdot (1+1) \cdot (\frac{w^2}{x^2} +1)] + [(x+d)^2 \cdot (\frac{2wx^2-[2xw^2]}{x^4})]$

$[(2)(x+d) \cdot (2) \cdot (\frac{w^2}{x^2} +1)] + [(x+d)^2 \cdot (\frac{2wx-2w^2}{x^3})]$

$[(4)(x+d) \cdot (\frac{w^2}{x^2} +1)] + [(x+d)^2 \cdot (\frac{2wx-2w^2}{x^3})]$

From this point, I could do stuff like factor out a $2(x+d)$, but it looks like mine won't be very easy to solve for zero.

This is what the derivative is supposed to be, which is easier to solve for zero:

$(x+d)^2(-\frac{2w^2}{x^3})+(\frac{w^2}{x^2}+1)(2)(x+d)$

What am I doing incorrectly?

Thanks!

2. ## Re: Trouble Finding Derivative

Originally Posted by Algebrah
I've been working on optimization problems and have been having problems with this one. The only thing that I'm confused about is what's going on when I find the derivative of the function:

$\frac{d}{dx}[(x+d)^2(\frac{w^2}{x^2} +1)]$

Also, in case it matters, $d=24$ and $w=6$, but in the example, the derivative is found with respect to the variables rather than their respective values.

My attempt:

$[(2)(x+d) \cdot (1+1) \cdot (\frac{w^2}{x^2} +1)] + [(x+d)^2 \cdot (\frac{2wx^2-[2xw^2]}{x^4})]$

$[(2)(x+d) \cdot (2) \cdot (\frac{w^2}{x^2} +1)] + [(x+d)^2 \cdot (\frac{2wx-2w^2}{x^3})]$

$[(4)(x+d) \cdot (\frac{w^2}{x^2} +1)] + [(x+d)^2 \cdot (\frac{2wx-2w^2}{x^3})]$

From this point, I could do stuff like factor out a $2(x+d)$, but it looks like mine won't be very easy to solve for zero.

This is what the derivative is supposed to be, which is easier to solve for zero:

$(x+d)^2(-\frac{2w^2}{x^3})+(\frac{w^2}{x^2}+1)(2)(x+d)$

What am I doing incorrectly?

Thanks!
1. You have introduced and extra factor of two in the first term erroneously.

2. The second term should be:

$(x+d)^2 \frac{d}{dx}\left( \frac{w^2}{x^2}+1\right) =(x+d)^2 \frac{(-2)w^2}{x^3}$

CB

3. ## Re: Trouble Finding Derivative

Originally Posted by Algebrah
I've been working on optimization problems and have been having problems with this one. The only thing that I'm confused about is what's going on when I find the derivative of the function:

$\frac{d}{dx}[(x+d)^2(\frac{w^2}{x^2} +1)]$

Also, in case it matters, $d=24$ and $w=6$, but in the example, the derivative is found with respect to the variables rather than their respective values.

My attempt:

$[(2)(x+d) \cdot (1+1) \cdot (\frac{w^2}{x^2} +1)] + [(x+d)^2 \cdot (\frac{2wx^2-[2xw^2]}{x^4})]$
Where did that "(1+1)"? Surely you didn't differentiate "(x+ d)" and get "(1+ 1)"? "d" is a constant. The derivative of x+ d with respect to x is 1. Also you appear to have tried to write $\frac{w^2}{x^2}+ 1= \frac{w^2+ x^2}{x^2}$ and use the quotient rule- but you have done that incorrectly. Better is to write $\frac{w^2}{x^2}+ 1= w^2x^{-2}+ 1$ so the derivative is $-2w^2x^{-3}= \frac{-2w^2}{x^3}$.

$[(2)(x+d) \cdot (2) \cdot (\frac{w^2}{x^2} +1)] + [(x+d)^2 \cdot (\frac{2wx-2w^2}{x^3})]$

$[(4)(x+d) \cdot (\frac{w^2}{x^2} +1)] + [(x+d)^2 \cdot (\frac{2wx-2w^2}{x^3})]$

From this point, I could do stuff like factor out a $2(x+d)$, but it looks like mine won't be very easy to solve for zero.

This is what the derivative is supposed to be, which is easier to solve for zero:

$(x+d)^2(-\frac{2w^2}{x^3})+(\frac{w^2}{x^2}+1)(2)(x+d)$

What am I doing incorrectly?

Thanks!

4. ## Re: Trouble Finding Derivative

Perhaps if I hadn't neglected the $\frac{d}{dx}$ in front, things would've come together a lot quicker! Lesson learned.

Thanks a lot for the help!