Differentiate fuction

**miss.strw** · October 15th 2009, 10:41 AM

Differentiate the following function.

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!

**Mush** · October 15th 2009, 10:47 AM

Originally Posted by miss.strw

Differentiate the following function.

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!

You have done it correctly. Although it is bad form to write it the way you've written it. You should have:

$y = \sqrt{x} (5x - 2)$
$y = x^{\frac{1}{2}} (5x - 2)$
$y = 5x^{\frac{3}{2}} - 2x^{\frac{1}{2}}$

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

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

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

Now write it with square root signs!

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

An easier way to do this might been with the product rule, which states that if $y = f(x) g(x)$ , then $y' = f'(x) g(x) + f(x) g'(x)$

**miss.strw** · October 15th 2009, 10:56 AM

Thank you so much!!!! thank you!!!

i didn't know it can go more than that!
n make it look "prettier" lol cool!!! thanks! learnt something!

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