What is (\partial)/(\partial y)(xcos(y-x)) ?

The answer to (\partial)/(\partial y)(xcos(y-x)) is -xsin(y-x)

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(\partial)/(\partial y)(xcos(y-x))

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

- x sin (y - x)

Solution steps

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

Treat

x

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the chain rule:

- sin (y - x) \frac{\partial}{\partial y} (y - x)

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

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

= x (- sin (y - x) \cdot 1)

Simplify

= - x sin (y - x)

\int \frac{3 x ^{2}}{x ^{3} + 7} d x

\frac{d}{d x} (x y)

y^{^{''}} + 2 y^{^{'}} + 2 y = e^{- t} sin (t)

f^{^{'}} (x) = (\frac{1 - sec ( x )}{tan ( x )})

\frac{\partial}{\partial x} (1 - x^{2})

What is (\partial)/(\partial y)(xcos(y-x)) ?
The answer to (\partial)/(\partial y)(xcos(y-x)) is -xsin(y-x)