What is (\partial)/(\partial x)(cos(3x+yz)*3) ?

The answer to (\partial)/(\partial x)(cos(3x+yz)*3) is -9sin(3x+yz)

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(\partial)/(\partial x)(cos(3x+yz)*3)

\frac{\partial}{\partial x} (cos (3 x + yz) \cdot 3)

- 9 sin (3 x + yz)

Solution steps

\frac{\partial}{\partial x} (cos (3 x + yz) \cdot 3)

Treat

y, z

as constants

Take the constant out:

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

= 3 \frac{\partial}{\partial x} (cos (3 x + yz))

Apply the chain rule:

- sin (3 x + yz) \frac{\partial}{\partial x} (3 x + yz)

= - sin (3 x + yz) \frac{\partial}{\partial x} (3 x + yz)

\frac{\partial}{\partial x} (3 x + yz) = 3

= 3 (- sin (3 x + yz) \cdot 3)

Simplify

= - 9 sin (3 x + yz)

\frac{d}{d x} ((\frac{x - 3}{x + 4}) (x^{2} + 2 x + 1))

\int_{0}^{3} sin^{2} (m x) d x

\frac{1}{6 x - 12} 2 x + 16

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

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

What is (\partial)/(\partial x)(cos(3x+yz)*3) ?
The answer to (\partial)/(\partial x)(cos(3x+yz)*3) is -9sin(3x+yz)