What is (\partial)/(\partial x)(3ycos(-3x)) ?

The answer to (\partial)/(\partial x)(3ycos(-3x)) is -9ysin(3x)

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(\partial)/(\partial x)(3ycos(-3x))

\frac{\partial}{\partial x} (3 y cos (- 3 x))

- 9 y sin (3 x)

Solution steps

\frac{\partial}{\partial x} (3 y cos (- 3 x))

Simplify

3 y cos (- 3 x) : 3 y cos (3 x)

= \frac{\partial}{\partial x} (3 y cos (3 x))

Treat

y

as a constant

Take the constant out:

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

= 3 y \frac{\partial}{\partial x} (cos (3 x))

Apply the chain rule:

- sin (3 x) \frac{\partial}{\partial x} (3 x)

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

\frac{\partial}{\partial x} (3 x) = 3

= 3 y (- sin (3 x) \cdot 3)

Simplify

= - 9 y sin (3 x)

\frac{d}{d x} (x^{2} - x - 12)

tan (7) x^{2}

\int_{0}^{1} 2 - x^{2} d x

y = \frac{x}{1 + x ^{2}}, (0, 0)

\frac{d}{d x} (\frac{1}{3} x^{3} + 1)

What is (\partial)/(\partial x)(3ycos(-3x)) ?
The answer to (\partial)/(\partial x)(3ycos(-3x)) is -9ysin(3x)