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)

\int \frac{1}{x + x x} d x

\int 8 x + \frac{1}{x} d x

\frac{d}{d x} (\frac{x + 4}{3 x ^{2} e ^{x}})

\int_{0}^{1} e^{x} d x

\frac{d}{d x} (x + a)

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