What is (\partial)/(\partial y)(x^2sin(4y)) ?

The answer to (\partial)/(\partial y)(x^2sin(4y)) is x^2cos(4y)*4

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(\partial)/(\partial y)(x^2sin(4y))

\frac{\partial}{\partial y} (x^{2} sin (4 y))

x^{2} cos (4 y) \cdot 4

Solution steps

\frac{\partial}{\partial y} (x^{2} sin (4 y))

Treat

x

as a constant

Take the constant out:

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

= x^{2} \frac{\partial}{\partial y} (sin (4 y))

Apply the chain rule:

cos (4 y) \frac{\partial}{\partial y} (4 y)

= cos (4 y) \frac{\partial}{\partial y} (4 y)

\frac{\partial}{\partial y} (4 y) = 4

= x^{2} cos (4 y) \cdot 4

\frac{\partial}{\partial z} (ln (x^{2} + y^{2}) - z)

n = 0 \sum \infty 2 n + 1

\int 39 sec^{4} (x) d x

y^{^{'}} = (t + y)^{2} - 1, y (2) = 5

f (x) = \frac{1}{5 - x}, at x = 0

What is (\partial)/(\partial y)(x^2sin(4y)) ?
The answer to (\partial)/(\partial y)(x^2sin(4y)) is x^2cos(4y)*4