How Does Taylor's Series Apply to Multivariable Functions?

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Are you serious? You have never seen Taylor's series before and your teacher assigned a problem like this? klWhat an evil person!

The Taylor's series, at (0,0), for a function, f(x,y), of two variables is given by
f(0,0)+ f_x(0,0)x+ f_y(0,0)y+ \frac{f_{xx}(0,0)}{2}x^2+ \frac{f_{xy}(0,0)}{2}xy+ \frac{f_{yy}(0,0)}{2}y^2+ \frac{f_{xxx}(0,0)}{6}x^3+ \frac{f_{xxy}(0,0)}{6}x^2y+ \frac{f_{xyy}(0,0)}{6}xy^2+ \frac{f_{yyy}(0,0)}{6}y^3+ \cdot\cdot\cdot

Can you find those derivatives?
 

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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