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

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

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

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

cos (x - y) + 1

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

= cos (x - y) + 1 - 0

Simplify

= cos (x - y) + 1

\int e^{\frac{x}{l}} d x

\int (x^{\frac{4}{3}} - 3 x^{\frac{5}{2}}) d x

\int_{0}^{2 π} sin^{2} (x co) s^{2} x

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

\frac{\partial}{\partial x} (y^{2} + 2 x^{2} - y)

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