What is the first derivative of 1+e^x ?

The first derivative of 1+e^x is e^x

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derivative of 1+e^x

derivative of

1 + e^{x}

e^{x}

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

= 0 + e^{x}

Simplify

= e^{x}

y^{^{''}} + 2 y^{^{'}} + y = 8 e^{- t}

\frac{d y}{d x} = - \frac{2 x y + y ^{2} + 1}{x ^{2} + 2 x y}

k = 1 \sum \infty \frac{2 ^{k}}{k !}

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