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

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

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

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

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

Solution steps

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

Treat

x

as a constant

Apply the chain rule:

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

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

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

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

\frac{d y}{d x} = \frac{3 y}{2 x}

\int ln (9) d x

\frac{d}{d x} ((4 x^{2} - 3) (4 x^{2} + 3))

(y^{2} + 1) d x = y sec^{2} (x) d y

\frac{d}{d z} (sinh (z + iπ))

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