What is (\partial)/(\partial y)(2ye^x-ln(z)) ?

The answer to (\partial)/(\partial y)(2ye^x-ln(z)) is 2e^x

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(\partial)/(\partial y)(2ye^x-ln(z))

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

2 e^{x}

Solution steps

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

Treat

x, z

as constants

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

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

\frac{\partial}{\partial y} (ln (z)) = 0

= 2 e^{x} - 0

Simplify

= 2 e^{x}

\int_{0}^{6} f (x) d x

\frac{d}{d x} (x^{2} + 2 x + 8)

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

f (x) = (2) (e^{3 x})

\frac{d}{d x} (- ∣ x ∣)

What is (\partial)/(\partial y)(2ye^x-ln(z)) ?
The answer to (\partial)/(\partial y)(2ye^x-ln(z)) is 2e^x