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

The answer to (\partial)/(\partial x)(sin(9x+yz)) is cos(9x+yz)*9

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(\partial)/(\partial x)(sin(9x+yz))

\frac{\partial}{\partial x} (sin (9 x + yz))

cos (9 x + yz) \cdot 9

Solution steps

\frac{\partial}{\partial x} (sin (9 x + yz))

Treat

y, z

as constants

Apply the chain rule:

cos (9 x + yz) \frac{\partial}{\partial x} (9 x + yz)

= cos (9 x + yz) \frac{\partial}{\partial x} (9 x + yz)

\frac{\partial}{\partial x} (9 x + yz) = 9

= cos (9 x + yz) \cdot 9

\frac{d}{d x} (\frac{x ^{3} - 2 x + 4}{x ^{3}})

\int x (5 x - 3)^{2} d x

\int e^{3 x} + \frac{1}{x} d x

\frac{d}{d x} (x \frac{d y}{d x})

y^{^{'}} + 3 y = 2 e^{- 3 t}

What is (\partial)/(\partial x)(sin(9x+yz)) ?
The answer to (\partial)/(\partial x)(sin(9x+yz)) is cos(9x+yz)*9