What is (\partial)/(\partial y)(2z^2sin(y)) ?

The answer to (\partial)/(\partial y)(2z^2sin(y)) is 2z^2cos(y)

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(\partial)/(\partial y)(2z^2sin(y))

\frac{\partial}{\partial y} (2 z^{2} sin (y))

2 z^{2} cos (y)

Solution steps

\frac{\partial}{\partial y} (2 z^{2} sin (y))

Treat

z

as a constant

Take the constant out:

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

= 2 z^{2} \frac{\partial}{\partial y} (sin (y))

Apply the common derivative:

\frac{\partial}{\partial y} (sin (y)) = cos (y)

= 2 z^{2} cos (y)

e^{s}

2 cos (7 x), sin (14 x)

500 q - \frac{q ^{2}}{ln ( q + 300 )}

(x + 2 y) d x + x d y = 0

x \to - 2 lim (1 + \frac{2}{x + 2})

What is (\partial)/(\partial y)(2z^2sin(y)) ?
The answer to (\partial)/(\partial y)(2z^2sin(y)) is 2z^2cos(y)