What is (\partial)/(\partial x)(4ycos(9x)) ?

The answer to (\partial)/(\partial x)(4ycos(9x)) is -36ysin(9x)

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(\partial)/(\partial x)(4ycos(9x))

\frac{\partial}{\partial x} (4 y cos (9 x))

- 36 y sin (9 x)

Solution steps

\frac{\partial}{\partial x} (4 y cos (9 x))

Treat

y

as a constant

Take the constant out:

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

= 4 y \frac{\partial}{\partial x} (cos (9 x))

Apply the chain rule:

- sin (9 x) \frac{\partial}{\partial x} (9 x)

= - sin (9 x) \frac{\partial}{\partial x} (9 x)

\frac{\partial}{\partial x} (9 x) = 9

= 4 y (- sin (9 x) \cdot 9)

Simplify

= - 36 y sin (9 x)

- x^{2} + 3, 2 x + 3

\int (\frac{2}{x} + 3 sin (x) - 4 e x) d x

\frac{d}{d x} (4 x - 8 y^{3})

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

\int_{0}^{1} (2 x + 1) e^{x^{2} + x} d x

What is (\partial)/(\partial x)(4ycos(9x)) ?
The answer to (\partial)/(\partial x)(4ycos(9x)) is -36ysin(9x)