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

The answer to (\partial)/(\partial x)(2sin(2x)cos(2y)) is 4cos(2x)cos(2y)

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

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

4 cos (2 x) cos (2 y)

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the chain rule:

cos (2 x) \frac{\partial}{\partial x} (2 x)

= cos (2 x) \frac{\partial}{\partial x} (2 x)

\frac{\partial}{\partial x} (2 x) = 2

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

Simplify

= 4 cos (2 x) cos (2 y)

\frac{\partial}{\partial x} (\frac{1}{x + 1})

\int (2 t^{2} - 1)^{2} d t

2 (x) + 1

- 7 x^{2} + 63, - x^{2} + 9

x \to - 1 lim (\frac{1 - x ^{2}}{x + 1})

What is (\partial)/(\partial x)(2sin(2x)cos(2y)) ?
The answer to (\partial)/(\partial x)(2sin(2x)cos(2y)) is 4cos(2x)cos(2y)