What is (\partial)/(\partial y)(sin(x-y)+(x-y)) ?

The answer to (\partial)/(\partial y)(sin(x-y)+(x-y)) is -cos(x-y)-1

Popular >

(\partial)/(\partial y)(sin(x-y)+(x-y))

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

- cos (x - y) - 1

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

= - cos (x - y) + 0 - 1

Simplify

= - cos (x - y) - 1

\frac{d}{d x} (x - 3 e^{x})

\int 2 sin (x) cos (x) d x

\frac{\partial}{\partial x} (ln (x^{2} + 3))

x \to 2 lim (\frac{2 x - 4}{x - 2})

(3 x^{2} + y - 4) - (2 y - x) \frac{d y}{d x} = 0

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