What is (\partial)/(\partial x)(-2y^3sin(2x)) ?

The answer to (\partial)/(\partial x)(-2y^3sin(2x)) is -4y^3cos(2x)

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(\partial)/(\partial x)(-2y^3sin(2x))

\frac{\partial}{\partial x} (- 2 y^{3} sin (2 x))

- 4 y^{3} cos (2 x)

Solution steps

\frac{\partial}{\partial x} (- 2 y^{3} sin (2 x))

Treat

y

as a constant

Take the constant out:

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

= - 2 y^{3} \frac{\partial}{\partial x} (sin (2 x))

Apply the chain rule:

cos (2 x) \frac{\partial}{\partial x} (2 x)

= cos (2 x) \frac{\partial}{\partial x} (2 x)

\frac{\partial}{\partial x} (2 x) = 2

= - 2 y^{3} cos (2 x) \cdot 2

Simplify

= - 4 y^{3} cos (2 x)

d x - x + y d y = 0

\int e^{17 x} d x

\frac{\partial}{\partial y} (3 x yz + x y^{2} - x z^{3})

42 y^{^{''}} + 29 y^{^{'}} - 5 y = 0

n = 1 \sum \infty \frac{n}{( 3 + n )}

What is (\partial)/(\partial x)(-2y^3sin(2x)) ?
The answer to (\partial)/(\partial x)(-2y^3sin(2x)) is -4y^3cos(2x)