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

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

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

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

sin (y) cos (2 x + y) \cdot 2

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the chain rule:

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

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

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

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

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\int (x + 3) (x^{2} + 6 x + 7)^{6} d x

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