What is (\partial)/(\partial y)(x^3-3x+y-e^y) ?

The answer to (\partial)/(\partial y)(x^3-3x+y-e^y) is -e^y+1

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(\partial)/(\partial y)(x^3-3x+y-e^y)

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

- e^{y} + 1

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial y} (x^{3}) = 0

\frac{\partial}{\partial y} (3 x) = 0

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

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

= 0 - 0 + 1 - e^{y}

Simplify

= - e^{y} + 1

x \to 9 lim (\frac{x ^{2} - 2}{x - 9})

5 t e^{- \frac{t}{7}}

\frac{\partial}{\partial x} (e^{- x^{8} - y^{4}})

\int 6 sin^{3} (6 x) cos^{3} (6 x) d x

x \to \infty + lim (3 e^{x})

What is (\partial)/(\partial y)(x^3-3x+y-e^y) ?
The answer to (\partial)/(\partial y)(x^3-3x+y-e^y) is -e^y+1