What is (\partial)/(\partial y)(8cos(ye^{xy})sin(x)) ?

The answer to (\partial)/(\partial y)(8cos(ye^{xy})sin(x)) is -8sin(e^{yx}y)sin(x)(e^{yx}+e^{yx}yx)

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(\partial)/(\partial y)(8cos(ye^{xy})sin(x))

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

- 8 sin (e^{y x} y) sin (x) (e^{y x} + e^{y x} y x)

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

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

Apply the chain rule:

- sin (y e^{x y}) \frac{\partial}{\partial y} (y e^{x y})

= - sin (y e^{x y}) \frac{\partial}{\partial y} (y e^{x y})

\frac{\partial}{\partial y} (y e^{x y}) = e^{x y} + x e^{x y} y

= 8 sin (x) (- sin (y e^{x y}) (e^{x y} + x e^{x y} y))

Simplify

= - 8 sin (e^{y x} y) sin (x) (e^{y x} + e^{y x} y x)

\int (sin (θ))^{3} d θ

4 sec (x)

x \to - 9 - lim (\frac{1}{x - 9})

x \to 0 + lim (ln (0))

x y^{^{'}} + y = 9 x y^{2}

What is (\partial)/(\partial y)(8cos(ye^{xy})sin(x)) ?
The answer to (\partial)/(\partial y)(8cos(ye^{xy})sin(x)) is -8sin(e^{yx}y)sin(x)(e^{yx}+e^{yx}yx)