What is (\partial)/(\partial x)(9sin(2x+y)) ?

The answer to (\partial)/(\partial x)(9sin(2x+y)) is 18cos(2x+y)

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(\partial)/(\partial x)(9sin(2x+y))

\frac{\partial}{\partial x} (9 sin (2 x + y))

18 cos (2 x + y)

Solution steps

\frac{\partial}{\partial x} (9 sin (2 x + y))

Treat

y

as a constant

Take the constant out:

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

= 9 \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

= 9 cos (2 x + y) \cdot 2

Simplify

= 18 cos (2 x + y)

\frac{\partial}{\partial x} (x^{2} y + y z^{3} + z^{5} x)

(6 y + 2 x - 7) d x + (4 y + 6 x - 1) d y = 0, y (- 1) = 2

\frac{d}{d z} ((x^{2} + y^{2} + z^{2})^{- \frac{1}{2}} + ln (x y z))

y^{^{'}} = 12 t^{2} e^{- y}

\int \frac{x ^{2}}{x ^{6} + 9} d x

What is (\partial)/(\partial x)(9sin(2x+y)) ?
The answer to (\partial)/(\partial x)(9sin(2x+y)) is 18cos(2x+y)