What is (\partial)/(\partial y)(xycos(yz)) ?

The answer to (\partial)/(\partial y)(xycos(yz)) is x(cos(yz)-yzsin(yz))

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(\partial)/(\partial y)(xycos(yz))

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

x (cos (yz) - yz sin (yz))

Solution steps

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

Treat

x, z

as constants

Take the constant out:

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

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

Apply the Product Rule:

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

f = y, g = cos (yz)

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

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

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

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

Simplify

= x (cos (yz) - yz sin (yz))

\frac{d y}{d x} = \frac{x ^{2} + 3}{y}

f (x) = ln (x^{2} - 4)

\int (x - 1) e^{- x} d x

\frac{d ^{2}}{d x ^{2}} (6 x + 17)

t \to \infty lim (- e^{- t})

What is (\partial)/(\partial y)(xycos(yz)) ?
The answer to (\partial)/(\partial y)(xycos(yz)) is x(cos(yz)-yzsin(yz))