# Differentiation help

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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cristo
Staff Emeritus
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:
$$\frac{d}{dx}x^{-8/5}=-\frac{8}{5}x^{-13/5}$$

cristo
Staff Emeritus
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.

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!