# Differentiate fuction

• Oct 15th 2009, 10:41 AM
miss.strw
Differentiate 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)

• Oct 15th 2009, 10:47 AM
Mush
Quote:

Originally Posted by miss.strw
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)

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 AM
miss.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!