Differentiate the following function.

http://i38.tinypic.com/2a99v5u.gif

y '= ?

ATTEMPT

y=x^1/2(5x-2)

= 5x^(3/2)-2x(1/2)

= (15/2)x^(1/2)-x^(-1/2)

Then from this step I am lost. So please help me out! THANKS! (Rofl)

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- Oct 15th 2009, 10:41 AMmiss.strwDifferentiate fuction
**Differentiate the following function.**

http://i38.tinypic.com/2a99v5u.gif

*y '*= ?

**ATTEMPT**

y=x^1/2(5x-2)

= 5x^(3/2)-2x(1/2)

= (15/2)x^(1/2)-x^(-1/2)

Then from this step I am lost. So please help me out! THANKS! (Rofl) - Oct 15th 2009, 10:47 AMMush
You have done it correctly. Although it is bad form to write it the way you've written it. You should have:

$\displaystyle y = \sqrt{x} (5x - 2) $

$\displaystyle y = x^{\frac{1}{2}} (5x - 2) $

$\displaystyle y = 5x^{\frac{3}{2}} - 2x^{\frac{1}{2}} $

$\displaystyle \therefore y' = \frac{15}{2} x^{\frac{1}{2}} - x^{\frac{-1}{2}} $

Now, you can make it look a bit prettier from here. Remember that $\displaystyle x^{-a} = \frac{1}{x^{a}} $. So we can write:

$\displaystyle y' = \frac{15}{2} x^{\frac{1}{2}} - \frac{1}{x^{\frac{1}{2}}} $

Now write it with square root signs!

$\displaystyle y' = \frac{15}{2} \sqrt{x} - \frac{1}{\sqrt{x}} $.

An easier way to do this might been with the product rule, which states that if $\displaystyle y = f(x) g(x) $, then $\displaystyle y' = f'(x) g(x) + f(x) g'(x) $ - Oct 15th 2009, 10:56 AMmiss.strw
Thank you so much!!!! thank you!!! (Rofl)(Talking)

i didn't know it can go more than that!

n make it look "prettier" lol cool!!! thanks! learnt something!