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

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

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

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

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

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

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

Apply the chain rule:

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

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

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

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

n = 0 \sum \infty 3 n - 1

\int \frac{12}{1 - cos ( 3 x )} d x

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

3^{2 x + 1}

\int \frac{1}{7 x - 2} d x

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