What is (\partial)/(\partial x)(y(e^x-1)) ?

The answer to (\partial)/(\partial x)(y(e^x-1)) is e^xy

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(\partial)/(\partial x)(y(e^x-1))

\frac{\partial}{\partial x} (y (e^{x} - 1))

e^{x} y

Solution steps

\frac{\partial}{\partial x} (y (e^{x} - 1))

Treat

y

as a constant

Take the constant out:

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

= y \frac{\partial}{\partial x} (e^{x} - 1)

Apply the Sum/Difference Rule:

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

= y (\frac{\partial}{\partial x} (e^{x}) - \frac{\partial}{\partial x} (1))

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

\frac{\partial}{\partial x} (1) = 0

= y (e^{x} - 0)

Simplify

= e^{x} y

x \to 0 lim (x^{2} + 16 x + 25)

k = 0 \sum \infty \frac{2}{k !}

\int (4 x - 3)^{2} d x

x \to 0 + lim ((ln (x))^{x})

\frac{d}{d x} (\frac{1}{x} ln (e^{x} + x))

What is (\partial)/(\partial x)(y(e^x-1)) ?
The answer to (\partial)/(\partial x)(y(e^x-1)) is e^xy