What is (\partial)/(\partial x)(x+ycos(x)) ?

The answer to (\partial)/(\partial x)(x+ycos(x)) is 1-ysin(x)

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(\partial)/(\partial x)(x+ycos(x))

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

1 - y sin (x)

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

= 1 - y sin (x)

\int \frac{sin ^{2} ( x )}{2} d x

\frac{d y}{d x}, y^{2} = x

\frac{\partial}{\partial y} (4 x^{2} y^{2})

f (x) = 6 x

cos (x) (\frac{d y}{d x}) - sin (x) y = 0

What is (\partial)/(\partial x)(x+ycos(x)) ?
The answer to (\partial)/(\partial x)(x+ycos(x)) is 1-ysin(x)