What is (\partial)/(\partial x)(3cos(3x+4y)) ?

The answer to (\partial)/(\partial x)(3cos(3x+4y)) is -9sin(3x+4y)

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(\partial)/(\partial x)(3cos(3x+4y))

\frac{\partial}{\partial x} (3 cos (3 x + 4 y))

- 9 sin (3 x + 4 y)

Solution steps

\frac{\partial}{\partial x} (3 cos (3 x + 4 y))

Treat

y

as a constant

Take the constant out:

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

= 3 \frac{\partial}{\partial x} (cos (3 x + 4 y))

Apply the chain rule:

- sin (3 x + 4 y) \frac{\partial}{\partial x} (3 x + 4 y)

= - sin (3 x + 4 y) \frac{\partial}{\partial x} (3 x + 4 y)

\frac{\partial}{\partial x} (3 x + 4 y) = 3

= 3 (- sin (3 x + 4 y) \cdot 3)

Simplify

= - 9 sin (3 x + 4 y)

\frac{d y}{d t} = \frac{t}{y - 2}, y (- 1) = 0

\frac{d}{d x} (\frac{3}{5 x ^{2}})

\int sin (x) sin (2 x) cos (3 x) d x

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

What is (\partial)/(\partial x)(3cos(3x+4y)) ?
The answer to (\partial)/(\partial x)(3cos(3x+4y)) is -9sin(3x+4y)