What is (\partial)/(\partial x)(e^{-2z}ysin(x)) ?

The answer to (\partial)/(\partial x)(e^{-2z}ysin(x)) is e^{-2z}ycos(x)

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(\partial)/(\partial x)(e^{-2z}ysin(x))

\frac{\partial}{\partial x} (e^{- 2 z} y sin (x))

e^{- 2 z} y cos (x)

Solution steps

\frac{\partial}{\partial x} (e^{- 2 z} y sin (x))

Treat

z, y

as constants

Take the constant out:

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

= e^{- 2 z} y \frac{\partial}{\partial x} (sin (x))

Apply the common derivative:

\frac{\partial}{\partial x} (sin (x)) = cos (x)

= e^{- 2 z} y cos (x)

cos (θ), 0.5

153 x^{2} - 0.06 x^{3}

y = (\frac{3}{5} x)^{4}

f (x) = ln (7 x)

\frac{\partial}{\partial z} (\frac{y}{1 + x ^{2} y ^{2}})

What is (\partial)/(\partial x)(e^{-2z}ysin(x)) ?
The answer to (\partial)/(\partial x)(e^{-2z}ysin(x)) is e^{-2z}ycos(x)