Posted by **Chelsea** on Friday, October 22, 2010 at 6:42pm.

So, I have a bunch of derivative problems for homework and the only resource I have is the textbook, since we don't have notes. I have just started and want to make sure I'm doing them correctly. Our teacher said we do not have to simplify them either, so I'm also confused if I'm stopping at the right point.

Differentiate.

1. Y(u)= (u^-2 + u^-3)(u^5 - 2u^2)

So, f(x)= (u^-2 + u^-3)

g(x)= (u^5 - 2u^2)

power rule:

= (-2u^-3 + -3u^-4)(5u^4 - 4u)

product rule:

= (-2u^-3 + -3u^-4) * D(5u^4 - 4u) + (5u^4 - 4u) * D(-2u^-3 + -3u^-4)

D(5u^4 - 4u)= 20u^3

D(-2u^-3 + -3u^-4)= 6u^-4 + 12u^-5

2. y = (x + 1)/(x^3 + x - 2)

y-prime=

[(x^3 + x - 2) * D(x + 1) - (x + 1) * D(x^3 + x - 2)]/(x^3 + x - 2)^2

D(x + 1)= x ?

D(x^3 + x - 2)= 4x^2 ?

Please help. I need to understand these, as I have plenty to do!

- Calc- Derivatives -
**Reiny**, Friday, October 22, 2010 at 6:51pm
1. My first-line derivative would be

Y' (u) = (u^-2 + u^-3)(5u^4 - 4u) + (u^5-2u^2)(-2u^-3 - 3u^-4)

If I were testing my students to see if they master the product rule, this is the line I would be looking for.

The rest is "window dressing", that is, simplifying it.

If you want a very simple explanation of the product rule, it would be ...

if y = (u)(v)

then y' = uv' + vu'

2. my first-line derivative of this quotient rule is

y' = [ (x^3+ x - 2)(1) - (x+1)(3x^2 + 1) ]/(x^3+x-2)^2

in short,

if y = u/v

then y' = (vu' - uv')/v^2

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