How do you differentiate $cos (y) - (x^{2} y^{3}) + 2 y = π$ $cos (y) -( x^2y^3) + 2y = pi$ ?

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Answer 1 · 2015-05-19T18:53:42

This is a bit of a vague question, as it can be differentiated in terms of either $x$ $x$ or $y$ $y$ or both $x and y$ $x and y$

I will assume for $x and y$ $x and y$

$\frac{d^{2}}{d x d y} cos (y) - (x^{2} y^{3}) + 2 y = \frac{d^{2}}{d x d y} π$ $d^2/(dxdy) cos(y) - (x^2y^3) + 2y = d^2/(dxdy) pi$

we then first differentiate over $y$ $y$ and treat $x$ $x$ as a constant

$\frac{d}{d x} - sin (y) - (3 x^{2} y^{2}) + 2 = \frac{d}{d x} 0$ $d/dx -sin(y) - (3x^2y^2) + 2 = d/dx 0$

then we can differentiate over $x$ $x$ and treat $y$ $y$ as a constant

$- 1 - (6 x y^{2}) = 0$ $-1 - (6xy^2) = 0$

$6 x y^{2} = - 1$ $6xy^2 = -1$

$x y^{2} = - \frac{1}{6}$ $xy^2 = -1/6$

How do you differentiate cos (y) -( x^2y^3) + 2y = picos(y)−(x2y3)+2y=πcos (y) -( x^2y^3) + 2y = pi?