Solving Derivative Problem using Chain/Product Rule

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I have a tricky derivative HW problem I'm working and am hoping someone might tell me if I'm doing this correctly or not. Thanks in advance!


Find g'(x) where g(x) = [(1+4x)^5] X [(-x^2+x+3)^8)]


By the product rule, I get:

g'x = [(d/dx((1+4x)^5)) X (-x^2+x+3)^8] + [(d/dx(-x^2+x+3)^8) X (1+4x)^5)]


Then, using the chain rule I get:

g'(x) = [((d/dx((1+4x)^5))(d/dx(1+4x))) X (-x^2+x+3)^8] + [((d/dx((-x^2+x+3)^8))(d/dx(-x^2+x+3))) X ((1+4x)^5)]


Giving:

g'(x) = [(20(1+4x)^4) X (-x^2+x+3)^8] + [(8(-x^2+x+3)^7 X (-2x+1)) X ((1+4x)^5)]


Then it should just be a matter of simplifying algebraically. Right?
 
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That's correct (even if it's a bit tough on the eyes :) )
 
Thanks for looking over that mess for me. I'll have to get used to using LaTeX I suppose. :)
 
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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