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Derivative of:
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Derivative of 5*tan(x)
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Identical expressions
y=ln(ctg2x^ three)
y equally ln(ctg2x cubed )
y equally ln(ctg2x to the power of three)
y=ln(ctg2x3)
y=lnctg2x3
y=ln(ctg2x³)
y=ln(ctg2x to the power of 3)
y=lnctg2x^3

The derivative of the function/ y=ln(ctg2x^3)

Derivative of y=ln(ctg2x^3)

The solution

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   /   3     \
log\cot (2*x)/

$$\log{\left(\cot^{3}{\left(2 x \right)} \right)}$$

log(cot(2*x)^3)

Detail solution

Let $u = \cot^{3}{\left(2 x \right)}$ .
The derivative of $\log{\left(u \right)}$ is $\frac{1}{u}$ .
Then, apply the chain rule. Multiply by $\frac{d}{d x} \cot^{3}{\left(2 x \right)}$ :
1. Let $u = \cot{\left(2 x \right)}$ .
2. Apply the power rule: $u^{3}$ goes to $3 u^{2}$
3. Then, apply the chain rule. Multiply by $\frac{d}{d x} \cot{\left(2 x \right)}$ :
  1. There are multiple ways to do this derivative.
    Method #1
    1. Rewrite the function to be differentiated:
      
      $\cot{\left(2 x \right)} = \frac{1}{\tan{\left(2 x \right)}}$
    2. Let $u = \tan{\left(2 x \right)}$ .
    3. Apply the power rule: $\frac{1}{u}$ goes to $- \frac{1}{u^{2}}$
    4. Then, apply the chain rule. Multiply by $\frac{d}{d x} \tan{\left(2 x \right)}$ :
      
      Rewrite the function to be differentiated:
      
      $\tan{\left(2 x \right)} = \frac{\sin{\left(2 x \right)}}{\cos{\left(2 x \right)}}$
      
      Apply the quotient rule, which is:
      
      $\frac{d}{d x} \frac{f{\left(x \right)}}{g{\left(x \right)}} = \frac{- f{\left(x \right)} \frac{d}{d x} g{\left(x \right)} + g{\left(x \right)} \frac{d}{d x} f{\left(x \right)}}{g^{2}{\left(x \right)}}$
      
      $f{\left(x \right)} = \sin{\left(2 x \right)}$ and $g{\left(x \right)} = \cos{\left(2 x \right)}$ .
      
      To find $\frac{d}{d x} f{\left(x \right)}$ :
      
      Let $u = 2 x$ .
      
      The derivative of sine is cosine:
      
      $\frac{d}{d u} \sin{\left(u \right)} = \cos{\left(u \right)}$
      
      Then, apply the chain rule. Multiply by $\frac{d}{d x} 2 x$ :
      
      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 of the chain rule is:
      
      $2 \cos{\left(2 x \right)}$
      
      To find $\frac{d}{d x} g{\left(x \right)}$ :
      
      Let $u = 2 x$ .
      
      The derivative of cosine is negative sine:
      
      $\frac{d}{d u} \cos{\left(u \right)} = - \sin{\left(u \right)}$
      
      Then, apply the chain rule. Multiply by $\frac{d}{d x} 2 x$ :
      
      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 of the chain rule is:
      
      $- 2 \sin{\left(2 x \right)}$
      
      Now plug in to the quotient rule:
      
      $\frac{2 \sin^{2}{\left(2 x \right)} + 2 \cos^{2}{\left(2 x \right)}}{\cos^{2}{\left(2 x \right)}}$
      
      The result of the chain rule is:
      
      $- \frac{2 \sin^{2}{\left(2 x \right)} + 2 \cos^{2}{\left(2 x \right)}}{\cos^{2}{\left(2 x \right)} \tan^{2}{\left(2 x \right)}}$
    Method #2
    1. Rewrite the function to be differentiated:
      
      $\cot{\left(2 x \right)} = \frac{\cos{\left(2 x \right)}}{\sin{\left(2 x \right)}}$
    2. Apply the quotient rule, which is:
      
      $\frac{d}{d x} \frac{f{\left(x \right)}}{g{\left(x \right)}} = \frac{- f{\left(x \right)} \frac{d}{d x} g{\left(x \right)} + g{\left(x \right)} \frac{d}{d x} f{\left(x \right)}}{g^{2}{\left(x \right)}}$
      
      $f{\left(x \right)} = \cos{\left(2 x \right)}$ and $g{\left(x \right)} = \sin{\left(2 x \right)}$ .
      
      To find $\frac{d}{d x} f{\left(x \right)}$ :
      
      Let $u = 2 x$ .
      
      The derivative of cosine is negative sine:
      
      $\frac{d}{d u} \cos{\left(u \right)} = - \sin{\left(u \right)}$
      
      Then, apply the chain rule. Multiply by $\frac{d}{d x} 2 x$ :
      
      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 of the chain rule is:
      
      $- 2 \sin{\left(2 x \right)}$
      
      To find $\frac{d}{d x} g{\left(x \right)}$ :
      
      Let $u = 2 x$ .
      
      The derivative of sine is cosine:
      
      $\frac{d}{d u} \sin{\left(u \right)} = \cos{\left(u \right)}$
      
      Then, apply the chain rule. Multiply by $\frac{d}{d x} 2 x$ :
      
      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 of the chain rule is:
      
      $2 \cos{\left(2 x \right)}$
      
      Now plug in to the quotient rule:
      
      $\frac{- 2 \sin^{2}{\left(2 x \right)} - 2 \cos^{2}{\left(2 x \right)}}{\sin^{2}{\left(2 x \right)}}$
  The result of the chain rule is:
  
  $- \frac{3 \left(2 \sin^{2}{\left(2 x \right)} + 2 \cos^{2}{\left(2 x \right)}\right) \cot^{2}{\left(2 x \right)}}{\cos^{2}{\left(2 x \right)} \tan^{2}{\left(2 x \right)}}$
The result of the chain rule is:

$- \frac{3 \left(2 \sin^{2}{\left(2 x \right)} + 2 \cos^{2}{\left(2 x \right)}\right)}{\cos^{2}{\left(2 x \right)} \tan^{2}{\left(2 x \right)} \cot{\left(2 x \right)}}$
Now simplify:

$- \frac{12}{\sin{\left(4 x \right)}}$

The answer is:

$- \frac{12}{\sin{\left(4 x \right)}}$

The graph

Plot the graph f(x)

Plot the graph f'(x)

The first derivative [src]

          2     
-6 - 6*cot (2*x)
----------------
    cot(2*x)

$$\frac{- 6 \cot^{2}{\left(2 x \right)} - 6}{\cot{\left(2 x \right)}}$$

Simplify

The second derivative [src]

   /                                 2\
   |                  /       2     \ |
   |         2        \1 + cot (2*x)/ |
12*|2 + 2*cot (2*x) - ----------------|
   |                        2         |
   \                     cot (2*x)    /

$$12 \left(- \frac{\left(\cot^{2}{\left(2 x \right)} + 1\right)^{2}}{\cot^{2}{\left(2 x \right)}} + 2 \cot^{2}{\left(2 x \right)} + 2\right)$$

Simplify

The third derivative [src]

                   /                             2                    \
                   |              /       2     \      /       2     \|
   /       2     \ |              \1 + cot (2*x)/    2*\1 + cot (2*x)/|
48*\1 + cot (2*x)/*|-2*cot(2*x) - ---------------- + -----------------|
                   |                    3                 cot(2*x)    |
                   \                 cot (2*x)                        /

$$48 \left(\cot^{2}{\left(2 x \right)} + 1\right) \left(- \frac{\left(\cot^{2}{\left(2 x \right)} + 1\right)^{2}}{\cot^{3}{\left(2 x \right)}} + \frac{2 \left(\cot^{2}{\left(2 x \right)} + 1\right)}{\cot{\left(2 x \right)}} - 2 \cot{\left(2 x \right)}\right)$$

Simplify

The graph

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Identical expressions

Derivative of y=ln(ctg2x^3)

The graph:

Piecewise:

The solution

Method #1

Method #2