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Derivative of:
Derivative of 4x^2
Derivative of 7
Derivative of coth(x)/tanh(x)
Derivative of 3*x
Identical expressions
(one +cos4x)/(one -cos4x)
(1 plus co sinus of e of 4x) divide by (1 minus co sinus of e of 4x)
(one plus co sinus of e of 4x) divide by (one minus co sinus of e of 4x)
1+cos4x/1-cos4x
(1+cos4x) divide by (1-cos4x)
Similar expressions
(1-cos4x)/(1-cos4x)
(1+cos4x)/(1+cos4x)

The derivative of the function/ (1+cos4x)/(1-cos4x)

Derivative of (1+cos4x)/(1-cos4x)

The solution

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1 + cos(4*x)
------------
1 - cos(4*x)

$$\frac{\cos{\left(4 x \right)} + 1}{1 - \cos{\left(4 x \right)}}$$

d /1 + cos(4*x)\
--|------------|
dx\1 - cos(4*x)/

$$\frac{d}{d x} \frac{\cos{\left(4 x \right)} + 1}{1 - \cos{\left(4 x \right)}}$$

Transform

Detail solution

Apply the quotient rule, which is:

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

$f{\left(x \right)} = \cos{\left(4 x \right)} + 1$ and $g{\left(x \right)} = 1 - \cos{\left(4 x \right)}$ .

To find $\frac{d}{d x} f{\left(x \right)}$ :
1. Differentiate $\cos{\left(4 x \right)} + 1$ term by term:
  1. The derivative of the constant $1$ is zero.
  2. Let $u = 4 x$ .
  3. The derivative of cosine is negative sine:
    
    $\frac{d}{d u} \cos{\left(u \right)} = - \sin{\left(u \right)}$
  4. Then, apply the chain rule. Multiply by $\frac{d}{d x} 4 x$ :
    1. 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: $4$
    The result of the chain rule is:
    
    $- 4 \sin{\left(4 x \right)}$
  The result is: $- 4 \sin{\left(4 x \right)}$
To find $\frac{d}{d x} g{\left(x \right)}$ :
1. Differentiate $1 - \cos{\left(4 x \right)}$ term by term:
  1. The derivative of the constant $1$ is zero.
  2. The derivative of a constant times a function is the constant times the derivative of the function.
    1. Let $u = 4 x$ .
    2. The derivative of cosine is negative sine:
      
      $\frac{d}{d u} \cos{\left(u \right)} = - \sin{\left(u \right)}$
    3. Then, apply the chain rule. Multiply by $\frac{d}{d x} 4 x$ :
      1. The derivative of a constant times a function is the constant times the derivative of the function.
        
        Apply the power rule: $x$ goes to $1$
        
        So, the result is: $4$
      The result of the chain rule is:
      
      $- 4 \sin{\left(4 x \right)}$
    So, the result is: $4 \sin{\left(4 x \right)}$
  The result is: $4 \sin{\left(4 x \right)}$
Now plug in to the quotient rule:

$\frac{- 4 \cdot \left(1 - \cos{\left(4 x \right)}\right) \sin{\left(4 x \right)} - 4 \left(\cos{\left(4 x \right)} + 1\right) \sin{\left(4 x \right)}}{\left(1 - \cos{\left(4 x \right)}\right)^{2}}$
Now simplify:

$- \frac{8 \sin{\left(4 x \right)}}{\left(\cos{\left(4 x \right)} - 1\right)^{2}}$

The answer is:

$- \frac{8 \sin{\left(4 x \right)}}{\left(\cos{\left(4 x \right)} - 1\right)^{2}}$

The graph

Plot the graph f(x)

Plot the graph f'(x)

The first derivative [src]

   4*sin(4*x)    4*(1 + cos(4*x))*sin(4*x)
- ------------ - -------------------------
  1 - cos(4*x)                      2     
                      (1 - cos(4*x))

$$- \frac{4 \sin{\left(4 x \right)}}{1 - \cos{\left(4 x \right)}} - \frac{4 \left(\cos{\left(4 x \right)} + 1\right) \sin{\left(4 x \right)}}{\left(1 - \cos{\left(4 x \right)}\right)^{2}}$$

Simplify

The second derivative [src]

   /                               /      2                 \           \
   |                               | 2*sin (4*x)            |           |
   |      2         (1 + cos(4*x))*|------------- + cos(4*x)|           |
   | 2*sin (4*x)                   \-1 + cos(4*x)           /           |
16*|------------- - ----------------------------------------- + cos(4*x)|
   \-1 + cos(4*x)                 -1 + cos(4*x)                         /
-------------------------------------------------------------------------
                              -1 + cos(4*x)

$$\frac{16 \left(\cos{\left(4 x \right)} - \frac{\left(\cos{\left(4 x \right)} + 1\right) \left(\cos{\left(4 x \right)} + \frac{2 \sin^{2}{\left(4 x \right)}}{\cos{\left(4 x \right)} - 1}\right)}{\cos{\left(4 x \right)} - 1} + \frac{2 \sin^{2}{\left(4 x \right)}}{\cos{\left(4 x \right)} - 1}\right)}{\cos{\left(4 x \right)} - 1}$$

Simplify

The third derivative [src]

   /                                                                   /                            2        \\         
   |       /      2                 \                                  |       6*cos(4*x)      6*sin (4*x)   ||         
   |       | 2*sin (4*x)            |                   (1 + cos(4*x))*|-1 + ------------- + ----------------||         
   |     3*|------------- + cos(4*x)|                                  |     -1 + cos(4*x)                  2||         
   |       \-1 + cos(4*x)           /     3*cos(4*x)                   \                     (-1 + cos(4*x)) /|         
64*|-1 + ---------------------------- + ------------- - ------------------------------------------------------|*sin(4*x)
   \            -1 + cos(4*x)           -1 + cos(4*x)                       -1 + cos(4*x)                     /         
------------------------------------------------------------------------------------------------------------------------
                                                     -1 + cos(4*x)

$$\frac{64 \left(-1 - \frac{\left(\cos{\left(4 x \right)} + 1\right) \left(-1 + \frac{6 \cos{\left(4 x \right)}}{\cos{\left(4 x \right)} - 1} + \frac{6 \sin^{2}{\left(4 x \right)}}{\left(\cos{\left(4 x \right)} - 1\right)^{2}}\right)}{\cos{\left(4 x \right)} - 1} + \frac{3 \left(\cos{\left(4 x \right)} + \frac{2 \sin^{2}{\left(4 x \right)}}{\cos{\left(4 x \right)} - 1}\right)}{\cos{\left(4 x \right)} - 1} + \frac{3 \cos{\left(4 x \right)}}{\cos{\left(4 x \right)} - 1}\right) \sin{\left(4 x \right)}}{\cos{\left(4 x \right)} - 1}$$

Simplify

The graph

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Derivative of (1+cos4x)/(1-cos4x)

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The solution