What is (\partial)/(\partial x)(-3ycos(-9x)) ?

The answer to (\partial)/(\partial x)(-3ycos(-9x)) is 27ysin(9x)

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(\partial)/(\partial x)(-3ycos(-9x))

\frac{\partial}{\partial x} (- 3 y cos (- 9 x))

27 y sin (9 x)

Solution steps

\frac{\partial}{\partial x} (- 3 y cos (- 9 x))

Simplify

- 3 y cos (- 9 x) : - 3 y cos (9 x)

= \frac{\partial}{\partial x} (- 3 y cos (9 x))

Treat

y

as a constant

Take the constant out:

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

= - 3 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

= - 3 y (- sin (9 x) \cdot 9)

Simplify

= 27 y sin (9 x)

f (x) = \frac{1}{r + 3 e ^{r}}

(8 + x) y^{^{'}} = 9 y

\frac{\partial}{\partial x} (y^{2} + 2 x y + 7 y + 7 x)

x \to π lim (c \cdot cos (x))

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

What is (\partial)/(\partial x)(-3ycos(-9x)) ?
The answer to (\partial)/(\partial x)(-3ycos(-9x)) is 27ysin(9x)