First Partial Derivative of f(x,y)=arctan (y/x)

teng125
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f(x,y)=arctan (y/x).
may i know what is the first partial derivative of this??

thanx
 
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Go back to your first calculus textbook and re-study the method of finding the derivative of Arctg(x).
 
teng125 said:
f(x,y)=arctan (y/x).
may i know what is the first partial derivative of this??

thanx

First partial derivative with respect to what? You have to know which variable you are taking the derivative with respect to when you try to find the partial derivative of a function of several variables.
 
If you wish simply to get answers, go to www.quickmath.com it will do derivatives, integrals etc., for the above problem use http://www.hostsrv.com/webmab/app1/MSP/quickmath/02/pageGenerate?site=quickmath&s1=calculus&s2=differentiate&s3=advanced .

Those who would help you here will require that you post some work (anything, an attempt, even if it is wrong) before they will respond.
 
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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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