Is my second derivative for y = (sinx)^2 correct?

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1. y = (sinx)^2



Homework Equations





The Attempt at a Solution



i found the 1st deriviative... 2(sinx)(cosx)

and then the second derivative... (-2(sinx)^2) + (cosx)^2

i don't have the answer. so i wanted to know if this was right.
 
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Not quite. Take a closer look. What's the derivative of 2(x)(x)? Using your method it would be 3x.
 

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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