What is the derivative of y=sin(x+y)?

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what is the derivative of y=sin(x+y)?

what is the derivative of y=sin(x+y)?
 
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The partial derivatives of f(x,y) = sin(x+y) are \frac{df}{dx} = cos(x+y) and \frac{df}{dy} = cos(x+y)
 
mattsoto - What do you want the derivative with respect to? If it's anything other than x or y, we need to talk a bit more.
 
the derivative is respect to y, Diane...
 
The derivative of y, given y= sin(x+ y), with respect to x, using implicit differentiation: y'= cos(x+y)(1+ y') so y'- y'cos(x+y)= cos(x+ y) and
y'= cos(x+y)/(1- cos(x+y)).

The derivative of x, given y= sin(x+ y), with respect to y (which is what you told you Diane you want, but I doubt since I would read what you originally wrote as 'the derivative OF y= ...), by implicit differentiation: 1= cos(x+y)(x'+ y) so
1- ycos(x+y)= cos(x+y)x' and x'= (1- ycos(x+y))/cos(x+y).
 

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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