Mastering Exponents: Simplifying and Applying Rules for Derivative Homework

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Homework Statement



([3√(x^2+4)^4]^2

Homework Equations



None needed.
Chain rule
product rule etc

The Attempt at a Solution


I stopped at:

[((x^2+4)^4)^1/3]^2

So I have 3 exponents. I don't know how to simplify this in order to move on to do the chain rule or whatever rule that comes next. The exponents are killing me
 
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Torshi said:

Homework Statement



([3√(x^2+4)^4]^2
Although you can't tell from the above, from your work below, it appears that the radical is a cube root.

Is this what you're trying to differentiate?
## (\sqrt[3]{(x^2 + 4)^4})^2##
Torshi said:

Homework Equations



None needed.
Chain rule
product rule etc

The Attempt at a Solution


I stopped at:

[((x^2+4)^4)^1/3]^2

So I have 3 exponents. I don't know how to simplify this in order to move on to do the chain rule or whatever rule that comes next. The exponents are killing me

What does (ar)s simplify to?
 
Mark44 said:
Although you can't tell from the above, from your work below, it appears that the radical is a cube root.

Is this what you're trying to differentiate?
## (\sqrt[3]{(x^2 + 4)^4})^2##What does (ar)s simplify to?

I think it simplified down to (x^2+4)^8/3
I multiplied the exponents: 1/3 * 4/1 * 2/1 = 8/3
 
OK, that's the first step.

Now, what is d/dx[(x2 + 4)8/3]?
 
Mark44 said:
OK, that's the first step.

Now, what is d/dx[(x2 + 4)8/3]?

I figured it out. Thank you. My main issue was with the exponents in regards to if I had to multiply all of them together which was true.
 
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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