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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(\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)

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

\int - \frac{1}{2} x^{3} e^{- x} d x

\int \frac{v}{1 + v ^{2}} d v

\int 0.06 q - 0.6 q - \frac{7700}{q ^{2}} d q

\frac{d y}{d x} = a - b 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)