What is (\partial)/(\partial z)(x^2yz^2+sin(yz)) ?

The answer to (\partial)/(\partial z)(x^2yz^2+sin(yz)) is 2x^2yz+cos(yz)y

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(\partial)/(\partial z)(x^2yz^2+sin(yz))

\frac{\partial}{\partial z} (x^{2} y z^{2} + sin (yz))

2 x^{2} yz + cos (yz) y

Solution steps

\frac{\partial}{\partial z} (x^{2} y z^{2} + sin (yz))

Treat

x, y

as constants

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial z} (x^{2} y z^{2}) + \frac{\partial}{\partial z} (sin (yz))

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

\frac{\partial}{\partial z} (sin (yz)) = cos (yz) y

= 2 x^{2} yz + cos (yz) y

\int 1 + \frac{x ^{2}}{2 - x ^{2}} d x

\int \frac{3 ^{x}}{2 ^{x}} d x

x \to 0 lim (15)

\int \frac{6 x}{x ^{2} + 4} d x

(- 2, 7), (0, 5)

What is (\partial)/(\partial z)(x^2yz^2+sin(yz)) ?
The answer to (\partial)/(\partial z)(x^2yz^2+sin(yz)) is 2x^2yz+cos(yz)y