What is (\partial)/(\partial x)(4e^{yz}cos(xyz)) ?

The answer to (\partial)/(\partial x)(4e^{yz}cos(xyz)) is -4e^{yz}yzsin(yzx)

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(\partial)/(\partial x)(4e^{yz}cos(xyz))

\frac{\partial}{\partial x} (4 e^{yz} cos (x yz))

- 4 e^{yz} yz sin (yz x)

Solution steps

\frac{\partial}{\partial x} (4 e^{yz} cos (x yz))

Treat

y, z

as constants

Take the constant out:

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

= 4 e^{yz} \frac{\partial}{\partial x} (cos (x yz))

Apply the chain rule:

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

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

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

= 4 e^{yz} (- sin (x yz) yz)

Simplify

= - 4 e^{yz} yz sin (yz x)

x \to 7 lim (\frac{- 2}{7 - x})

y = \frac{6 x}{x + 2}, (4, 4)

x \to 2 lim (x + 3)

\int \frac{( x - 2 )}{x ^{2} - 7 x + 12} d x

\int \frac{1}{x lo g _{10} ( x )} d x

What is (\partial)/(\partial x)(4e^{yz}cos(xyz)) ?
The answer to (\partial)/(\partial x)(4e^{yz}cos(xyz)) is -4e^{yz}yzsin(yzx)