Deriving a root and a fraction

Twinflower
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Homework Statement



Derive y= \sqrt(2x^4) - \frac{5}{3x^2}



The Attempt at a Solution



I am still at the first part of the function (the root):
First I tried to derive inside the root like this:
\sqrt(2x^4) = \sqrt(8x^3) = 2 \sqrt(2) \times \sqrt(x^3) = \sqrt(2) \times x^2

Unfortunately, the first part of the function is supposed to be 2\sqrt(2)x.

The second part I am nowhere close yet.
 
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Hi Twinflower! :smile:

Before you take the derivative, you should first try to simplify your expression.

Did you know that \sqrt {(2x)} = \sqrt 2 \sqrt x?
And that \sqrt {(x^4)} = x^2?
 
Last edited:
Hi, thanks for your reply.

I tried what you are suggesting, but I somehow ended up with \sqrt(x^4) = x^3 since \sqrt(x^2) = x^1 = x

Second attempt:

\sqrt(2x^4)
\sqrt(2) \times \sqrt(x^4)
\sqrt(2) \times x^2
2\times \sqrt(2) \times x
2 x

hm... didn't go quite well now either?
 
Actually, your second attempt is looking quite well. :)But how did you get from:

2\times \sqrt(2) \times x

to

2 x?
 
Oh, my bad.
Brain-fart.

For a second I though that 2 times sqrt2 equals 2.

The final answer should ofcourse be
2 \times \sqrt(2) \times x
 
Thank you by the way.

Now I need a push in the right direction regarding the fraction :)
 
Twinflower said:
Thank you by the way.

Now I need a push in the right direction regarding the fraction :)

You're welcome! :smile:For the fraction you need to know that for instance {1 \over x^5} = x^{-5}.

Oh, and also that {2 \over 3x} = {2\over 3} \times {1 \over x}.
 
Ah, of course!
Gimme a minute, and I'll figure that one out as well :)
 
Ok, here we go:


- \frac{5}{3x^2}
- \frac{5}{3} \times \frac{1}{x^2}
- \frac{5}{3} \times x^{-2}
- \frac{5}{3} \times -2 \times x^{-3}
\frac{10}{3} \times x^{-3}
\frac{10\times x^{-3}}{3}
 
  • #10
Thanks a bunch! :)
 
  • #11
Hey! Your minute is up! :wink:

Edit: errrr... I guess you were just in time! :rolleyes:
 
  • #12
yeye, I needed to write it down in my exercise paper and THEN i had to write it all in LaTeX (which is still quite messy to me)

:)
 
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