Quick question: Derivative of a Trig Function

momogiri
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Find the derivative of the function. Simplify if possible.
y = sin-1(2x + 3)

I'm wondering if the derivative is as simple as \frac{1}{\sqrt{1-(2x+3)^{2}}} XD
 
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you forgot to multiply by (2x+3)'.

(arcsinu)'=\frac{u'}{\sqrt{1-u^2}}
 
OMFG BOB YOU'RE BACCKKKKK <3333 XD
My assignment is already overdue, so I can't do anything now XD

So it'd be
\frac{(2x+3)&#039;}{\sqrt{1-(2x+3)^{2}}} to solve?
 
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Indeed, and it will hardly make the answer less simple :smile:
 
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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