What is (\partial)/(\partial x)(x^2+y^2-ycos(x)) ?

The answer to (\partial)/(\partial x)(x^2+y^2-ycos(x)) is 2x+ysin(x)

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(\partial)/(\partial x)(x^2+y^2-ycos(x))

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

2 x + y sin (x)

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial x} (x^{2}) = 2 x

\frac{\partial}{\partial x} (y^{2}) = 0

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

= 2 x + 0 - (- y sin (x))

Simplify

= 2 x + y sin (x)

\int_{1}^{2} \frac{1}{8 ( 2 + 2 x - x ^{2} )} d x

\int - \frac{x ^{2}}{( - x + 1 ) ( x + 1 )} d x

\frac{d}{d x} (A e^{x} + B e^{- x})

x y^{^{'}} = ((1 - x) y + x^{2} + x^{3}), y (3) = 2

x \to 3 - lim (\frac{x}{∣ x ∣})

What is (\partial)/(\partial x)(x^2+y^2-ycos(x)) ?
The answer to (\partial)/(\partial x)(x^2+y^2-ycos(x)) is 2x+ysin(x)