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

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

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

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

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

Solution steps

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

Treat

x, y

as constants

Take the constant out:

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

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

Apply the Product Rule:

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

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

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

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

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

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

Simplify

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

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

\frac{d y}{d x} = 2 y (1 - y)

\frac{x ^{2} + 5 x + 6}{x ^{2} - 4}

\frac{2}{( 6 x ^{2} + 5 x + 1 ) ^{2}}

\int \frac{1}{aa uu} d u

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