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

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

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

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

- y sin (x y)

Solution steps

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

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

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

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

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

Simplify

= - y sin (x y)

y^{^{'}} = - 7 y ln (\frac{10}{y})

\frac{d}{d x} (x^{2} + 6 x - 16)

\frac{\partial}{\partial h} (π r^{2} h)

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

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

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