derivative of (e^x)/(x+1) — step-by-step solution

1) Let's differentiate:

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

We differentiate

p1

e^{x}

and

q2

x + 1

using:

\frac{d}{d x} (\frac{p1}{q2}) = \frac{\frac{d}{d x} (p1) \cdot q2 - p1 \cdot \frac{d}{d x} (q2)}{{q2}^{2}}

2) Let's perform the differentiation

p1

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

ⓘ The derivative of exp(x) is exp(x):

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

e^{x}

e^{x}

3) Let's perform the differentiation

q2

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

ⓘ The derivative of a sum is the sum of the derivatives:

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

1 + 0

1 + 0

ⓘ Adding 0 changes nothing:

1 + 0

1

1

4) We evaluate the resulting derivative (

\frac{p1}{q2}

)':

\frac{e^{x} \cdot (x + 1) - e^{x} \cdot 1}{{(x + 1)}^{2}}

\frac{e^{x} \cdot (x + 1) - e^{x}}{{(x + 1)}^{2}}

ⓘ What the terms have in common is factored out, so that the fraction can be simplified:

\frac{e^{x} \cdot (x + 1) - e^{x}}{{(x + 1)}^{2}}

\frac{e^{x} \cdot ((x + 1) - 1)}{{(x + 1)}^{2}}

\frac{e^{x} \cdot ((x + 1) - 1)}{{(x + 1)}^{2}}

ⓘ Parentheses are computed first:

1 - 1

0

\frac{e^{x} \cdot (x + 0)}{{(x + 1)}^{2}}

ⓘ Adding 0 changes nothing:

x + 0

x

\frac{e^{x} x}{{(x + 1)}^{2}}

5) Then you can continue the differentiation:

ⅰ

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

\frac{e^{x} x}{{(x + 1)}^{2}}

▶Derivative:

\frac{e^{x} x}{{(x + 1)}^{2}}