What is the first derivative of e^x-4 ?

The first derivative of e^x-4 is e^x

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derivative of e^x-4

\frac{d}{d x} (e^{x} - 4)

e^{x}

Solution steps

\frac{d}{d x} (e^{x} - 4)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (e^{x}) - \frac{d}{d x} (4)

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

\frac{d}{d x} (4) = 0

= e^{x} - 0

Simplify

= e^{x}

\int \frac{x ^{2} + 2 x + 5}{( x + 1 ) ( x ^{2} + 1 )} d x

\int_{0}^{1} \frac{12}{x ^{2} - 4} d x

\frac{d}{d x} (\frac{1}{5 + x})

\frac{64}{16 + x ^{2}}, 1

\frac{\partial}{\partial x} (e^{3 x} cos (5 y))