What is (\partial)/(\partial y)(sin(x+2y+3z)) ?

The answer to (\partial)/(\partial y)(sin(x+2y+3z)) is cos(x+2y+3z)*2

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(\partial)/(\partial y)(sin(x+2y+3z))

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

cos (x + 2 y + 3 z) \cdot 2

Solution steps

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

Treat

x, z

as constants

Apply the chain rule:

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

= cos (x + 2 y + 3 z) \frac{\partial}{\partial y} (x + 2 y + 3 z)

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

= cos (x + 2 y + 3 z) \cdot 2

\int 2 cos^{3} (3 x) d x

\int_{0}^{\frac{π}{2}} 28 x^{2} sin (x) d x

\frac{d}{d x} (1 - x^{3})

f (x) = (2 x - 5)^{3}

(- 3.5) (6.2)

What is (\partial)/(\partial y)(sin(x+2y+3z)) ?
The answer to (\partial)/(\partial y)(sin(x+2y+3z)) is cos(x+2y+3z)*2