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

The answer to (\partial)/(\partial x)(ycos(x-y)) is -ysin(x-y)

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

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

- y sin (x - y)

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the chain rule:

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

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

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

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

Simplify

= - y sin (x - y)

64 y^{^{''}} + 80 y^{^{'}} + 25 y = 0, y (0) = a, y^{^{'}} (0) = - 5

f (x) = ln (x^{2} - 9 x + 1), at x = 9

\int e^{- x - y} d x

\int 7 arcsin (x) d x

\frac{d}{d x} ((\frac{x ^{2} + 3}{x ^{2} - 3})^{5})

What is (\partial)/(\partial x)(ycos(x-y)) ?
The answer to (\partial)/(\partial x)(ycos(x-y)) is -ysin(x-y)