What is (\partial)/(\partial x)(xy-y+sin(1)) ?

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

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(\partial)/(\partial x)(xy-y+sin(1))

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

y

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

\frac{\partial}{\partial x} (sin (1)) = 0

= y - 0 + 0

Simplify

= y

\int_{0}^{2} 2^{- θ} d θ

\int sin^{3} (4 x) cos (4 x) d x

\frac{\partial}{\partial x} (x^{- 1} y^{2} + x y^{2} + 5 x y)

\int_{1}^{16} \frac{1}{8 - x} d x

\frac{d y}{d x} + 2 x y = y + 4 x - 2

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