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

The answer to (\partial)/(\partial x)(x^3-3x+y-e^y) is 3x^2-3

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

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

3 x^{2} - 3

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial x} (x^{3}) = 3 x^{2}

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

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

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

= 3 x^{2} - 3 + 0 - 0

Simplify

= 3 x^{2} - 3

\frac{d ^{4}}{d x ^{4}} (8 x)

\frac{\partial}{\partial x} (\frac{3 x - 2 y}{5 x + y})

\frac{d}{d x} (- sin (x) - cos (x))

\frac{d}{d x} (- 2 x^{2} y)

y^{^{'}} = y (x y^{4} + 1)

What is (\partial)/(\partial x)(x^3-3x+y-e^y) ?
The answer to (\partial)/(\partial x)(x^3-3x+y-e^y) is 3x^2-3