What is (\partial)/(\partial x)(40xy^{-4}z) ?

The answer to (\partial)/(\partial x)(40xy^{-4}z) is (40z)/(y^4)

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(\partial)/(\partial x)(40xy^{-4}z)

\frac{\partial}{\partial x} (40 x y^{- 4} z)

\frac{40 z}{y ^{4}}

Solution steps

\frac{\partial}{\partial x} (40 x y^{- 4} z)

Treat

y, z

as constants

Take the constant out:

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

= 40 y^{- 4} z \frac{\partial}{\partial x} (x)

Apply the common derivative:

\frac{\partial}{\partial x} (x) = 1

= 40 y^{- 4} z \cdot 1

Simplify

40 y^{- 4} z \cdot 1 : \frac{40 z}{y ^{4}}

= \frac{40 z}{y ^{4}}

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\frac{d}{d x} ((5 x - 3)^{4} (2 x - 7)^{- 3})

(\frac{1}{2 x})^{^{'}}

θ cos (θ)

\int (2 x - 4^{x}) d x

What is (\partial)/(\partial x)(40xy^{-4}z) ?
The answer to (\partial)/(\partial x)(40xy^{-4}z) is (40z)/(y^4)