Need help Determining the Derivative

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f(x)= -3x^4/(4x-8)^(1/2),

f'(x)=?

Cannot figure out how to do this one, would appreciate help finding my mistakes!I'm sure you all cna find the answer yourself but for convenience it is -3x^3(7x-16)/(4(x-2)^(3/2))
 

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Your answer is right, but we need to somehow get x-2 instead of 4x-8:

At the step:

\frac{(4x-8)(-12x^3) +6x^4}{(4x-8)^{\frac{1}{2}}(4x-8)}

We can factor a number out of (4x-8)1/2. We can also factor a number out of (4x-8). Once you get to the last step of your work, you can factor out a number from (14x-32).
 
i got it! thanks a lot i just didn't see the equivalency 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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