Friday

March 6, 2015

March 6, 2015

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

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 6:51pm1. 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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