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

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

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

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

- sin (x - y) + 1

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

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

Simplify

= - sin (x - y) + 1

y = x, y = 2 x 25 - x^{2}

\int_{- 2}^{2} - 2 x^{2} + 8 d x

f (x) = \frac{- 2}{x ^{2}}

\frac{1}{x ^{2} + 18 x + 88}

y = x (x - 4)

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