What is (\partial)/(\partial z)(-9cos(3x+yz)z) ?

The answer to (\partial)/(\partial z)(-9cos(3x+yz)z) is -9(-yzsin(3x+yz)+cos(3x+yz))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

(\partial)/(\partial z)(-9cos(3x+yz)z)

\frac{\partial}{\partial z} (- 9 cos (3 x + yz) z)

- 9 (- yz sin (3 x + yz) + cos (3 x + yz))

Solution steps

\frac{\partial}{\partial z} (- 9 cos (3 x + yz) z)

Treat

x, y

as constants

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= - 9 \frac{\partial}{\partial z} (cos (3 x + yz) z)

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = cos (3 x + yz), g = z

= - 9 (\frac{\partial}{\partial z} (cos (3 x + yz)) z + \frac{\partial}{\partial z} (z) cos (3 x + yz))

\frac{\partial}{\partial z} (cos (3 x + yz)) = - sin (3 x + yz) y

\frac{\partial}{\partial z} (z) = 1

= - 9 ((- sin (3 x + yz) y) z + 1 \cdot cos (3 x + yz))

Simplify

= - 9 (- yz sin (3 x + yz) + cos (3 x + yz))

What is (\partial)/(\partial z)(-9cos(3x+yz)z) ?
The answer to (\partial)/(\partial z)(-9cos(3x+yz)z) is -9(-yzsin(3x+yz)+cos(3x+yz))