What is (\partial)/(\partial y)(cos(8x)sin(4y)) ?

The answer to (\partial)/(\partial y)(cos(8x)sin(4y)) is cos(8x)cos(4y)*4

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(\partial)/(\partial y)(cos(8x)sin(4y))

\frac{\partial}{\partial y} (cos (8 x) sin (4 y))

cos (8 x) cos (4 y) \cdot 4

Solution steps

\frac{\partial}{\partial y} (cos (8 x) sin (4 y))

Treat

x

as a constant

Take the constant out:

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

= cos (8 x) \frac{\partial}{\partial y} (sin (4 y))

Apply the chain rule:

cos (4 y) \frac{\partial}{\partial y} (4 y)

= cos (4 y) \frac{\partial}{\partial y} (4 y)

\frac{\partial}{\partial y} (4 y) = 4

= cos (8 x) cos (4 y) \cdot 4

e^{- 12 t} f + 12 e^{- 12 t} f^{^{'}} = - 4 e^{- 14 t}

6 y^{^{''}} + 18 y^{^{'}} - 24 y = 0

y = \frac{1}{4 x ^{2}}, y = x, y = \frac{x}{16}

\int (12 x + 14) (3 x^{2} + 7 x - 1)^{5} d x

n = 1 \sum \infty \frac{6}{n} - \frac{6}{n + 1}

What is (\partial)/(\partial y)(cos(8x)sin(4y)) ?
The answer to (\partial)/(\partial y)(cos(8x)sin(4y)) is cos(8x)cos(4y)*4