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

The answer to (\partial)/(\partial x)(1-xycos(piy)) is -ycos(piy)

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

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

- y cos (π y)

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

= 0 - y cos (π y)

Simplify

= - y cos (π y)

\int \frac{1}{x ^{2} 64 - x ^{2}} d x

y^{^{''}} - 2 y^{^{'}} = e^{x}

ma c l a u r in \frac{1}{( 1 + x ) ^{2}}

\frac{d}{d x} (e^{2 x - 1} ln (2 x - 1))

\int \frac{9}{( x ^{2} - 1 ) ^{2}} d x

What is (\partial)/(\partial x)(1-xycos(piy)) ?
The answer to (\partial)/(\partial x)(1-xycos(piy)) is -ycos(piy)