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Graphing y =:
e^(2(x+1))/(2(x+1))
Identical expressions
e^(two (x+ one))/(two (x+ one))
e to the power of (2(x plus 1)) divide by (2(x plus 1))
e to the power of (two (x plus one)) divide by (two (x plus one))
e(2(x+1))/(2(x+1))
e2x+1/2x+1
e^2x+1/2x+1
e^(2(x+1)) divide by (2(x+1))
Similar expressions
e^(2(x-1))/(2(x+1))
e^(2(x+1))/(2(x-1))

The derivative of the function/ e^(2(x+1))/(2(x+1))

Derivative of e^(2(x+1))/(2(x+1))

The solution

You have entered [src]

 2*(x + 1)
E         
----------
2*(x + 1)

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

E^(2*(x + 1))/((2*(x + 1)))

Detail solution

Apply the quotient rule, which is:

$\frac{d}{d x} \frac{f{\left(x \right)}}{g{\left(x \right)}} = \frac{- f{\left(x \right)} \frac{d}{d x} g{\left(x \right)} + g{\left(x \right)} \frac{d}{d x} f{\left(x \right)}}{g^{2}{\left(x \right)}}$

$f{\left(x \right)} = e^{2 \left(x + 1\right)}$ and $g{\left(x \right)} = 2 x + 2$ .

To find $\frac{d}{d x} f{\left(x \right)}$ :
1. Let $u = 2 \left(x + 1\right)$ .
2. The derivative of $e^{u}$ is itself.
3. Then, apply the chain rule. Multiply by $\frac{d}{d x} 2 \left(x + 1\right)$ :
  1. The derivative of a constant times a function is the constant times the derivative of the function.
    1. Differentiate $x + 1$ term by term:
      1. Apply the power rule: $x$ goes to $1$
      2. The derivative of the constant $1$ is zero.
      The result is: $1$
    So, the result is: $2$
  The result of the chain rule is:
  
  $2 e^{2 x + 2}$
To find $\frac{d}{d x} g{\left(x \right)}$ :
1. Differentiate $2 x + 2$ term by term:
  1. The derivative of the constant $2$ is zero.
  2. The derivative of a constant times a function is the constant times the derivative of the function.
    1. Apply the power rule: $x$ goes to $1$
    So, the result is: $2$
  The result is: $2$
Now plug in to the quotient rule:

$\frac{2 \left(2 x + 2\right) e^{2 x + 2} - 2 e^{2 x + 2}}{\left(2 x + 2\right)^{2}}$
Now simplify:

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

The answer is:

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

The graph

Plot the graph f(x)

Plot the graph f'(x)

The first derivative [src]

                         2 + 2*x 
      1      2 + 2*x    e        
2*---------*e        - ----------
  2*(x + 1)                     2
                       2*(x + 1)

$$2 \frac{1}{2 \left(x + 1\right)} e^{2 x + 2} - \frac{e^{2 x + 2}}{2 \left(x + 1\right)^{2}}$$

Simplify

The second derivative [src]

/       1         2  \  2 + 2*x
|2 + -------- - -----|*e       
|           2   1 + x|         
\    (1 + x)         /         
-------------------------------
             1 + x

$$\frac{\left(2 - \frac{2}{x + 1} + \frac{1}{\left(x + 1\right)^{2}}\right) e^{2 x + 2}}{x + 1}$$

Simplify

The third derivative [src]

/      6        3          6    \  2 + 2*x
|4 - ----- - -------- + --------|*e       
|    1 + x          3          2|         
\            (1 + x)    (1 + x) /         
------------------------------------------
                  1 + x

$$\frac{\left(4 - \frac{6}{x + 1} + \frac{6}{\left(x + 1\right)^{2}} - \frac{3}{\left(x + 1\right)^{3}}\right) e^{2 x + 2}}{x + 1}$$

Simplify

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Derivative of e^(2(x+1))/(2(x+1))

The graph:

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The solution