What is (\partial)/(\partial y)(1-xycos(piy)) ?

The answer to (\partial)/(\partial y)(1-xycos(piy)) is -x(-piysin(piy)+cos(piy))

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(\partial)/(\partial y)(1-xycos(piy))

\frac{\partial}{\partial y} (1 - x y cos (π y))

- x (- π y sin (π y) + cos (π y))

Solution steps

\frac{\partial}{\partial y} (1 - x y cos (π y))

Treat

x

as a constant

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{\partial}{\partial y} (1) - \frac{\partial}{\partial y} (x y cos (π y))

\frac{\partial}{\partial y} (1) = 0

\frac{\partial}{\partial y} (x y cos (π y)) = x (cos (π y) - π y sin (π y))

= 0 - x (cos (π y) - π y sin (π y))

Simplify

= - x (- π y sin (π y) + cos (π y))

- ln (1 - x)

x \to - \infty lim (\frac{x}{2})

\frac{d y}{d x} - tan (x) y = y^{3}, y (0) = 1

\int 5 x^{4} e^{x^{5}} d x

\frac{d}{d t} (t^{3} + t)

What is (\partial)/(\partial y)(1-xycos(piy)) ?
The answer to (\partial)/(\partial y)(1-xycos(piy)) is -x(-piysin(piy)+cos(piy))