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))

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(\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))

n = 1 \sum \infty n e^{- 5 n}

cos^{4} (θ + 9)

y = x^{4} + 4 e^{x}, (0, 4)

\frac{d}{d x} ((\frac{1 - 2 x}{1 + x})^{3})

\frac{d}{d x} (\frac{4 x}{7 - tan ( x )})

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))