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

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(\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 y}{d x} = 6 y e^{x + 4}

t^{2} + 2 t - 3

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

n = 1 \sum \infty \frac{2}{3 n + 9}

\int_{l n (x)}^{1} sin^{3} (u) d u

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