What is (\partial)/(\partial y)(cos(2x)sin(3y)) ?

The answer to (\partial)/(\partial y)(cos(2x)sin(3y)) is cos(2x)cos(3y)*3

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(\partial)/(\partial y)(cos(2x)sin(3y))

\frac{\partial}{\partial y} (cos (2 x) sin (3 y))

cos (2 x) cos (3 y) \cdot 3

Solution steps

\frac{\partial}{\partial y} (cos (2 x) sin (3 y))

Treat

x

as a constant

Take the constant out:

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

= cos (2 x) \frac{\partial}{\partial y} (sin (3 y))

Apply the chain rule:

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

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

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

= cos (2 x) cos (3 y) \cdot 3

25 y^{^{''}} - 40 y^{^{'}} + 16 y = 0, y (4) = 2, y^{^{'}} (4) = 2

\frac{d}{d x} (tan (x - 5))

\int xg (x^{2}) d x

\int \frac{1}{5 x + 1} d x

\int \frac{8 x ^{2} - 5 x + 1}{x ^{2} + x} d x

What is (\partial)/(\partial y)(cos(2x)sin(3y)) ?
The answer to (\partial)/(\partial y)(cos(2x)sin(3y)) is cos(2x)cos(3y)*3