1. Derivatives

Hi!

I just started learning calculus this semester and it's a total brainbreak. We are on derivatives now and we just had a quiz about derivatives and limits. The problem is, when I thought I already knew derivatives I felt stupid during the whole 3 hours of answering the quiz. I found out I couldn't work on the complicated equations (especially the ones with the bar lines and radical signs) using the method our instructor taught us. I ended up having answered only one item which was a limit problem. Then a senior told me that I should have used the theorems for derivatives (not sure how to call that).

Here's one from our quiz:
$\displaystyle y=(1+3x-x^2)(x^2-5)$

$\displaystyle \frac {dy}{dx} = (x^2-5) \frac {d}{dx} (1+3x-x^2) + (1+3x-x^2) \frac {d}{dx} (x^2-5)$

$\displaystyle \frac {dy}{dx} = (x^2-5) [ \frac {d}{dx} (1) + \frac {d}{dx}(3x) + \frac {d}{dx} (x^2) ] + (1+3x-x^2) [ \frac {d}{dx} (x^2) - \frac {d}{dx}(5) ]$

$\displaystyle \frac {dy}{dx} = (x^2-5) (3+2x) + (1+3x-x^2) (2x)$

$\displaystyle \frac {dy}{dx} = 2x^3 + 3x^2 - 10x - 15 - 2x^3 + 6x^2 +2x$

$\displaystyle \frac {dy}{dx} = 9x^2 - 8x - 15$
Is my answer correct? Unfortunately I can no longer get any mark for getting it right so it's good as incorrect.

2. that's the way its done except in your first term it should be (x^2-5)*(3-2x) not 3+2x

3. Originally Posted by Xerxes

Here's one from our quiz:
$\displaystyle y=(1+3x-x^2)(x^2-5)$

$\displaystyle \frac {dy}{dx} = (x^2-5) \frac {d}{dx} (1+3x-x^2) + (1+3x-x^2) \frac {d}{dx} (x^2-5)$

$\displaystyle \frac {dy}{dx} = (x^2-5) [ \frac {d}{dx} (1) + \frac {d}{dx}(3x) + \frac {d}{dx} (x^2) ] + (1+3x-x^2) [ \frac {d}{dx} (x^2) - \frac {d}{dx}(5) ]$

$\displaystyle \frac {dy}{dx} = (x^2-5) (3+2x) + (1+3x-x^2) (2x)$
In those lines there's something wrong, figure out by yourself.

4. Originally Posted by Krizalid
In those lines there's something wrong, figure out by yourself.
Oh! Is there?
Okay, maybe I'll check it by morning. I have to get some rest now. Thanks a lot.
By the way, is there anything wrong about my syntax?

5. I see nothing wrong in your syntax.

6. crap!

Originally Posted by Calculus26
that's the way its done except in your first term it should be (x^2-5)*(3-2x) not 3+2x
Oh no! I did it again!
That's not fair. I can't screw up in the exams because of signs.

$\displaystyle \frac {dy}{dx} = (x^2-5) [ \frac {d}{dx} (1) + \frac {d}{dx}(3x) - \frac {d}{dx} (x^2) ] + (1+3x-x^2) [ \frac {d}{dx} (x^2) - \frac {d}{dx}(5) ]$

$\displaystyle \frac {dy}{dx} = (x^2-5) (3-2x) + (1+3x-x^2) (2x)$

$\displaystyle \frac {dy}{dx} = 3x^2 - 2x^3 + 10x - 15 - 2x^3 + 6x^2 +2x$

$\displaystyle \frac {dy}{dx} = 9x^2 + 12x - 15 - 4x^3$

Now, how do I thank you two, electronically? (EDIT: I see. I just have to click the right button.)

7. Originally Posted by Xerxes
Oh no! I did it again!
That's not fair. I can't screw up in the exams because of signs.

$\displaystyle \frac {dy}{dx} = (x^2-5) [ \frac {d}{dx} (1) + \frac {d}{dx}(3x) - \frac {d}{dx} (x^2) ] + (1+3x-x^2) [ \frac {d}{dx} (x^2) - \frac {d}{dx}(5) ]$

$\displaystyle \frac {dy}{dx} = (x^2-5) (3-2x) + (1+3x-x^2) (2x)$

$\displaystyle \frac {dy}{dx} = 3x^2 - 2x^3 + 10x - 15 - 2x^3 + 6x^2 +2x$

$\displaystyle \frac {dy}{dx} = 9x^2 + 12x - 15 - 4x^3$

Now, how do I thank you two, electronically? (EDIT: I see. I just have to click the right button.)
Or slide twenty dollar bills into that little slot under your monitor!