Differentiation Help - Find the du/dx Answer Here

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Awesomesauce
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Could someone help me with this simple differentiation?
u= 5root(1/x^8)
--> =(1/x^8)^1/5
= (1/5)/(x^8/5)
du/dx= ?
I don't know what to do after here, as I keep getting the wrong answer.
The answer is du/dx = (-8/5)x^-13/5
:( Thanks!
 
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Let's clarify the question. Do you want to differentiate

u=\Big({\frac{1}{x^8}}\Big)^{1/5}=x^{-8/5}

What's the derivative of x^n wrt x?
 
Yes. I want to differentiate that. How did you get to x^-8/5? I would know how to go from there, as its just reducing the power by one, and mutiplying the coefficient (1) by -8/5.
Thankyou for moving my question by the way! You have been really helpful so far!
 
Two rules:

\left(x^a\right)^b=x^{ab}

x^{-a}=\frac{1}{x^a}

Putting these together with your problem:
\left(\frac{1}{x^8}\right)^{1/5}=\left(x^{-8}\right)^{1/5}=x^{-8\times(1/5)}=x^{-8/5}.

Now, using the differentiation rule as you have already worked out:
\frac{d}{dx}x^{-8/5}=-\frac{8}{5}x^{-13/5}
 
Awesomesauce said:
Yes. I want to differentiate that. How did you get to x^-8/5?

Sorry, I should've been more explicit there. See ubiquitousuk's detailed post.

Thankyou for moving my question by the way! You have been really helpful so far!

You're welcome.
 
ubiquitousuk said:
Two rules:

\left(x^a\right)^b=x^{ab}

x^{-a}=\frac{1}{x^a}

Putting these together with your problem:
\left(\frac{1}{x^8}\right)^{1/5}=\left(x^{-8}\right)^{1/5}=x^{-8\times(1/5)}=x^{-8/5}.

Now, using the differentiation rule as you have already worked out:
\frac{d}{dx}x^{-8/5}=-\frac{8}{5}x^{-13/5}

Ah! Seems so simple now! Really helpful! Thanks!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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