Solve Chain Rule Equation: y=(tan^-1(6x))^2

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Homework Statement


y = (tan^-1(6x))^2

Homework Equations


Chain Rule, power rule?

The Attempt at a Solution


Okay, so I did power rule to bring it to 2(tan^-1(6x))
Then, I know to use the chain rule...
I get 2(tan^-1(6x)*(1/1+x^2)... I know u = 6x so I play 6 into x^2 and I get 6x^2...
I see everyone else has gotten 36x^2 on the bottom and I have no idea why, can someone please explain this problem IN DETAIL to me because for all the chain rule problems relevant to one like this, it just skips the algebraic steps which I need to know in this...
 
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I found out how they got 36x^2, but please a detailed review of this problem would still help!
 
robren said:

Homework Statement


y = (tan^-1(6x))^2

Homework Equations


Chain Rule, power rule?

The Attempt at a Solution


Okay, so I did power rule to bring it to 2(tan^-1(6x))
Then, I know to use the chain rule...
I get 2(tan^-1(6x)*(1/1+x^2)... I know u = 6x so I play 6 into x^2 and I get 6x^2...
I see everyone else has gotten 36x^2 on the bottom and I have no idea why, can someone please explain this problem IN DETAIL to me because for all the chain rule problems relevant to one like this, it just skips the algebraic steps which I need to know in this...
y' = 2 tan-1(6x) * d/dx(tan-1(6x))

For that last derivative, imagine that it is d/dx(tan-1(u)). What would that be? You need the chain rule for this derivative.
 
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Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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