Derivative of tg((2x-4)/x)

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   /2*x - 4\
tan|-------|
   \   x   /

\tan{\left(\frac{2 x - 4}{x} \right)}

tan((2*x - 4)/x)

Detail solution

Rewrite the function to be differentiated:

$\tan{\left(\frac{2 x - 4}{x} \right)} = \frac{\sin{\left(\frac{2 x - 4}{x} \right)}}{\cos{\left(\frac{2 x - 4}{x} \right)}}$
Apply the quotient rule, which is:

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

$f{\left(x \right)} = \sin{\left(\frac{2 x - 4}{x} \right)}$ and $g{\left(x \right)} = \cos{\left(\frac{2 x - 4}{x} \right)}$ .

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

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

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

The answer is:

\frac{4}{x^{2} \cos^{2}{\left(2 - \frac{4}{x} \right)}}

The graph

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The first derivative [src]

/       2/2*x - 4\\ /2   2*x - 4\
|1 + tan |-------||*|- - -------|
\        \   x   // |x       2  |
                    \       x   /

\left(\frac{2}{x} - \frac{2 x - 4}{x^{2}}\right) \left(\tan^{2}{\left(\frac{2 x - 4}{x} \right)} + 1\right)

Simplify

The second derivative [src]

  /       2/2*(-2 + x)\\ /    -2 + x\ /       /    -2 + x\    /2*(-2 + x)\\
4*|1 + tan |----------||*|1 - ------|*|-1 + 2*|1 - ------|*tan|----------||
  \        \    x     // \      x   / \       \      x   /    \    x     //
---------------------------------------------------------------------------
                                      2                                    
                                     x

\frac{4 \left(1 - \frac{x - 2}{x}\right) \left(2 \left(1 - \frac{x - 2}{x}\right) \tan{\left(\frac{2 \left(x - 2\right)}{x} \right)} - 1\right) \left(\tan^{2}{\left(\frac{2 \left(x - 2\right)}{x} \right)} + 1\right)}{x^{2}}

Simplify

The third derivative [src]

                                      /                                                    2                                        2                 \
  /       2/2*(-2 + x)\\ /    -2 + x\ |       /    -2 + x\    /2*(-2 + x)\     /    -2 + x\  /       2/2*(-2 + x)\\     /    -2 + x\     2/2*(-2 + x)\|
4*|1 + tan |----------||*|1 - ------|*|3 - 12*|1 - ------|*tan|----------| + 4*|1 - ------| *|1 + tan |----------|| + 8*|1 - ------| *tan |----------||
  \        \    x     // \      x   / \       \      x   /    \    x     /     \      x   /  \        \    x     //     \      x   /      \    x     //
-------------------------------------------------------------------------------------------------------------------------------------------------------
                                                                            3                                                                          
                                                                           x

\frac{4 \left(1 - \frac{x - 2}{x}\right) \left(\tan^{2}{\left(\frac{2 \left(x - 2\right)}{x} \right)} + 1\right) \left(4 \left(1 - \frac{x - 2}{x}\right)^{2} \left(\tan^{2}{\left(\frac{2 \left(x - 2\right)}{x} \right)} + 1\right) + 8 \left(1 - \frac{x - 2}{x}\right)^{2} \tan^{2}{\left(\frac{2 \left(x - 2\right)}{x} \right)} - 12 \left(1 - \frac{x - 2}{x}\right) \tan{\left(\frac{2 \left(x - 2\right)}{x} \right)} + 3\right)}{x^{3}}

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Derivative of tg((2x-4)/x)

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