What is the first derivative of e^x+e ?

The first derivative of e^x+e is e^x

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derivative of e^x+e

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

e^{x}

Solution steps

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

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (e^{x}) + \frac{d e}{d x}

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

\frac{d e}{d x} = 0

= e^{x} + 0

Simplify

= e^{x}

\int \frac{sin ^{2} ( x )}{cos ^{3} ( x )} d x

y = \frac{26}{5} - \frac{1}{x}, y = x

2 x^{2}, 2 x + 6, - 2, - 1

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

x \to 8 + lim (\frac{2}{x - 8})