What is (\partial)/(\partial y)(-9sin(3x+yz)) ?

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

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(\partial)/(\partial y)(-9sin(3x+yz))

\frac{\partial}{\partial y} (- 9 sin (3 x + yz))

- 9 cos (3 x + yz) z

Solution steps

\frac{\partial}{\partial y} (- 9 sin (3 x + yz))

Treat

x, z

as constants

Take the constant out:

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

= - 9 \frac{\partial}{\partial y} (sin (3 x + yz))

Apply the chain rule:

cos (3 x + yz) \frac{\partial}{\partial y} (3 x + yz)

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

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

= - 9 cos (3 x + yz) z

x \to \frac{π}{8} lim (tan (24 x))

tan (8 x)

\int_{0}^{1} 9 \frac{e ^{\frac{1}{x}}}{x ^{3}} d x

\frac{\partial}{\partial y} (8 e^{yz} cos (x yz))

(4.8) (2.12)

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