C2 Indices help

Given that y = 2x root(x) + 2 root(x)
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x

Show that dy/dx = x-1
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x root(x)

Any help?
Thanks

Simplify y to: $\displaystyle y = \frac{2x\sqrt{x} + 2\sqrt{x}}{x} = \frac{2x \sqrt{2}}{x} + \frac{2\sqrt{x}}{x} = 2\sqrt{x} + \frac{2}{\sqrt{x}} = 2x^{1/2} + 2x^{-1/2}$ then differentiate.

So after I differentiate I get x^-1/2 - x ^-3/2.
How do I simplify that to x-1 / x root(x)

Many thanks

So you have: $\displaystyle \frac{1}{\sqrt{x}} - \frac{1}{x\sqrt{x}} = \frac{x}{x\sqrt{x}} - \frac{1}{x\sqrt{x}} = \cdots$

Remember that: $x^{3/2} = x \cdot x^{1/2} = x \sqrt{x}$

Thanks a lot. Also i'm strugglng with:

root(x) - 1 / x^2

Find gradient when x = 2.
I get that i need to differentiate and then sub in 2 as x but no matter how many times I do it I can't get the right answer.

Again, first thing first - simplify, then differentiate:

$\displaystyle \frac{\sqrt{x} - 1}{x^2} = \frac{\sqrt{x}}{x^2} - \frac{1}{x^2} = x^{-3/2} - x^{-2}$

What do you get when you differentiate that?

-3/2x +2x^-3

See bolded bit. You need to subtract one from the power. What is $\displaystyle x^{-3/2 - 1}$?

What's $\displaystyle -\frac{3}{2} - 1 = -1.5 - 1$?

-3/2x^-5/2 + 2x^-3

And what do you get when you put $x=2$ into that?

-3/2(2^-5/2) + 2(2^-3) = -0.0152

That's correct, in't it?

Yes it is! Thank you so so much!