What is the derivative of (e^xdx) ?

The derivative of (e^xdx) is d(e^xx+e^x)

What is the first derivative of (e^xdx) ?

The first derivative of (e^xdx) is d(e^xx+e^x)

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

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

Solution steps

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

Treat

d

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = e^{x}, g = x

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

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

\frac{d x}{d x} = 1

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

Simplify

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

\frac{1}{s ^{4} - 9}

\frac{d}{d x} (\frac{( x - 4 ) ( x ^{2} + 4 x + 16 )}{x ^{3}})

\int \frac{1}{e ^{x} + 4} d x

x \to 2 lim (x^{3} - 5 x^{2} + 4)

\int \frac{3 x ^{2}}{16 - x ^{2}} d x

What is the derivative of (e^xdx) ?
The derivative of (e^xdx) is d(e^xx+e^x)
What is the first derivative of (e^xdx) ?
The first derivative of (e^xdx) is d(e^xx+e^x)