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Derivative of:
Derivative of -1/x
Derivative of x^(1/3)
Derivative of x*atan(x)
Derivative of (x^3)/3
Differential equation:
y'
Identical expressions
y'=(ctg*4x/ eleven)'
y stroke first (1st) order equally (ctg multiply by 4x divide by 11) stroke first (1st) order
y stroke first (1st) order equally (ctg multiply by 4x divide by eleven) stroke first (1st) order
y'=(ctg4x/11)'
y'=ctg4x/11'
y'=(ctg*4x divide by 11)'

The derivative of the function/ y'=(ctg*4x/11)'

Derivative of y'=(ctg*4x/11)'

The solution

You have entered [src]

cot(4*x)
--------
   11

$$\frac{\cot{\left(4 x \right)}}{11}$$

cot(4*x)/11

Detail solution

The derivative of a constant times a function is the constant times the derivative of the function.
1. There are multiple ways to do this derivative.
  Method #1
  1. Rewrite the function to be differentiated:
    
    $\cot{\left(4 x \right)} = \frac{1}{\tan{\left(4 x \right)}}$
  2. Let $u = \tan{\left(4 x \right)}$ .
  3. Apply the power rule: $\frac{1}{u}$ goes to $- \frac{1}{u^{2}}$
  4. Then, apply the chain rule. Multiply by $\frac{d}{d x} \tan{\left(4 x \right)}$ :
    1. Rewrite the function to be differentiated:
      
      $\tan{\left(4 x \right)} = \frac{\sin{\left(4 x \right)}}{\cos{\left(4 x \right)}}$
    2. Apply the quotient rule, which is:
      
      $\frac{d}{d x} \frac{f{\left(x \right)}}{g{\left(x \right)}} = \frac{- f{\left(x \right)} \frac{d}{d x} g{\left(x \right)} + g{\left(x \right)} \frac{d}{d x} f{\left(x \right)}}{g^{2}{\left(x \right)}}$
      
      $f{\left(x \right)} = \sin{\left(4 x \right)}$ and $g{\left(x \right)} = \cos{\left(4 x \right)}$ .
      
      To find $\frac{d}{d x} f{\left(x \right)}$ :
      
      Let $u = 4 x$ .
      
      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 $\frac{d}{d x} 4 x$ :
      
      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: $4$
      
      The result of the chain rule is:
      
      $4 \cos{\left(4 x \right)}$
      
      To find $\frac{d}{d x} g{\left(x \right)}$ :
      
      Let $u = 4 x$ .
      
      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 $\frac{d}{d x} 4 x$ :
      
      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: $4$
      
      The result of the chain rule is:
      
      $- 4 \sin{\left(4 x \right)}$
      
      Now plug in to the quotient rule:
      
      $\frac{4 \sin^{2}{\left(4 x \right)} + 4 \cos^{2}{\left(4 x \right)}}{\cos^{2}{\left(4 x \right)}}$
    The result of the chain rule is:
    
    $- \frac{4 \sin^{2}{\left(4 x \right)} + 4 \cos^{2}{\left(4 x \right)}}{\cos^{2}{\left(4 x \right)} \tan^{2}{\left(4 x \right)}}$
  Method #2
  1. Rewrite the function to be differentiated:
    
    $\cot{\left(4 x \right)} = \frac{\cos{\left(4 x \right)}}{\sin{\left(4 x \right)}}$
  2. Apply the quotient rule, which is:
    
    $\frac{d}{d x} \frac{f{\left(x \right)}}{g{\left(x \right)}} = \frac{- f{\left(x \right)} \frac{d}{d x} g{\left(x \right)} + g{\left(x \right)} \frac{d}{d x} f{\left(x \right)}}{g^{2}{\left(x \right)}}$
    
    $f{\left(x \right)} = \cos{\left(4 x \right)}$ and $g{\left(x \right)} = \sin{\left(4 x \right)}$ .
    
    To find $\frac{d}{d x} f{\left(x \right)}$ :
    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 $\frac{d}{d x} 4 x$ :
      
      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:
      
      $- 4 \sin{\left(4 x \right)}$
    To find $\frac{d}{d x} g{\left(x \right)}$ :
    1. Let $u = 4 x$ .
    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 $\frac{d}{d x} 4 x$ :
      
      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: $4$
      
      The result of the chain rule is:
      
      $4 \cos{\left(4 x \right)}$
    Now plug in to the quotient rule:
    
    $\frac{- 4 \sin^{2}{\left(4 x \right)} - 4 \cos^{2}{\left(4 x \right)}}{\sin^{2}{\left(4 x \right)}}$
So, the result is: $- \frac{4 \sin^{2}{\left(4 x \right)} + 4 \cos^{2}{\left(4 x \right)}}{11 \cos^{2}{\left(4 x \right)} \tan^{2}{\left(4 x \right)}}$
Now simplify:

$- \frac{4}{11 \sin^{2}{\left(4 x \right)}}$

The answer is:

$- \frac{4}{11 \sin^{2}{\left(4 x \right)}}$

The graph

Plot the graph f(x)

Plot the graph f'(x)

The first derivative [src]

            2     
  4    4*cot (4*x)
- -- - -----------
  11        11

$$- \frac{4 \cot^{2}{\left(4 x \right)}}{11} - \frac{4}{11}$$

Simplify

The second derivative [src]

   /       2     \         
32*\1 + cot (4*x)/*cot(4*x)
---------------------------
             11

$$\frac{32 \left(\cot^{2}{\left(4 x \right)} + 1\right) \cot{\left(4 x \right)}}{11}$$

Simplify

The third derivative [src]

     /       2     \ /         2     \
-128*\1 + cot (4*x)/*\1 + 3*cot (4*x)/
--------------------------------------
                  11

$$- \frac{128 \left(\cot^{2}{\left(4 x \right)} + 1\right) \left(3 \cot^{2}{\left(4 x \right)} + 1\right)}{11}$$

Simplify

The graph

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Identical expressions

Derivative of y'=(ctg*4x/11)'

The graph:

Piecewise:

The solution

Method #1

Method #2