What is (\partial)/(\partial x)(2ysin(10x-y)) ?

The answer to (\partial)/(\partial x)(2ysin(10x-y)) is 20ycos(10x-y)

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(\partial)/(\partial x)(2ysin(10x-y))

\frac{\partial}{\partial x} (2 y sin (10 x - y))

20 y cos (10 x - y)

Solution steps

\frac{\partial}{\partial x} (2 y sin (10 x - y))

Treat

y

as a constant

Take the constant out:

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

= 2 y \frac{\partial}{\partial x} (sin (10 x - y))

Apply the chain rule:

cos (10 x - y) \frac{\partial}{\partial x} (10 x - y)

= cos (10 x - y) \frac{\partial}{\partial x} (10 x - y)

\frac{\partial}{\partial x} (10 x - y) = 10

= 2 y cos (10 x - y) \cdot 10

Simplify

= 20 y cos (10 x - y)

What is (\partial)/(\partial x)(2ysin(10x-y)) ?
The answer to (\partial)/(\partial x)(2ysin(10x-y)) is 20ycos(10x-y)