Help! (1 Viewer)

professorayton8 · Sep 3, 2012

Hi, I really need help for these differentiation questions! 5/3(x)^1/2 and 2x^5+7x-1 all over 2x^2 Thank you!!

Appleorange · Sep 3, 2012

By "and" do you mean multiplication?

professorayton8 · Sep 3, 2012

No they're 2 separate questions

Appleorange · Sep 3, 2012

Oh. Miss read it.

One second.

Appleorange · Sep 3, 2012

I don't use Latex much
but

$<a href="http://www.codecogs.com/eqnedit.php?latex=(\frac{5x}{3})^{1/2}= \frac{1}{2}(\frac{5}{3})x^{-1/2}=\frac{5}{6}x^{-1/2}^{}= -\sqrt{\frac{5x}{3}}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?(\frac{5x}{3})^{1/2}= \frac{1}{2}(\frac{5}{3})x^{-1/2}=\frac{5}{6}x^{-1/2}^{}= -\sqrt{\frac{5x}{3}}" title="(\frac{5x}{3})^{1/2}= \frac{1}{2}(\frac{5}{3})x^{-1/2}=\frac{5}{6}x^{-1/2}^{}= -\sqrt{\frac{5x}{3}}" /></a>$

Appleorange · Sep 3, 2012

professorayton8 · Sep 3, 2012

Sorry, instead of the long question.... How do you do this question? X-2 over square root x?

professorayton8 · Sep 3, 2012

Thanks

Sy123 · Sep 3, 2012

professorayton8 said:
Sorry, instead of the long question.... How do you do this question? X-2 over square root x?

$y=\frac{x-2}{x}$

$u=x-2 \ \ u'=1 \ \ \ v=x \ \ v'=1 \\ y'=\frac{1(x)-1(x-2)}{x^2} \\ y=\frac{2}{x^2}$

Another way (faster)

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

Appleorange · Sep 3, 2012

I am tired, so I don't know how to embed this (It isn't fully simplified.

professorayton8 · Sep 3, 2012

The quotient rule! Smart

thanks

professorayton8 · Sep 3, 2012

The quotient rule! Smart

thanks

(ps it was square root

)

Appleorange · Sep 3, 2012

Appleorange said:
View attachment 26284

I am tired, so I don't know how to embed this (It isn't fully simplified.

Typed x'v-v'y

The x was meant to be a u

Appleorange · Sep 3, 2012

The other one

Appleorange · Sep 3, 2012

$\frac{2x^{5}+ 7x -1}{2x^{2}}= \frac{u'v-v'u}{v^{2}}=\frac{2x^{2}(10x^{4}+7)-4x(2x^{5}+7x-1)}{(2x^{2})^{2}}$

Success!

professorayton8 · Sep 3, 2012

Thanks

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