What is (\partial)/(\partial x)(-3xy^2-2ycos(x)) ?

The answer to (\partial)/(\partial x)(-3xy^2-2ycos(x)) is -3y^2+2ysin(x)

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(\partial)/(\partial x)(-3xy^2-2ycos(x))

\frac{\partial}{\partial x} (- 3 x y^{2} - 2 y cos (x))

- 3 y^{2} + 2 y sin (x)

Solution steps

\frac{\partial}{\partial x} (- 3 x y^{2} - 2 y cos (x))

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= - \frac{\partial}{\partial x} (3 x y^{2}) - \frac{\partial}{\partial x} (2 y cos (x))

\frac{\partial}{\partial x} (3 x y^{2}) = 3 y^{2}

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

= - 3 y^{2} - (- 2 y sin (x))

Simplify

= - 3 y^{2} + 2 y sin (x)

\int \frac{1}{t ^{2} 169 - t ^{2}} d t

y = \frac{x}{x + 6}, (4, 0.2)

\int_{1}^{e} ln (12 x) d x

y = x^{2}, y = x^{3}

x y^{^{'}} + y = - 8 x y^{2}

What is (\partial)/(\partial x)(-3xy^2-2ycos(x)) ?
The answer to (\partial)/(\partial x)(-3xy^2-2ycos(x)) is -3y^2+2ysin(x)