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Identical expressions
(4sin(x)+ nine)/(one +ctg^2x)
(4 sinus of (x) plus 9) divide by (1 plus ctg squared x)
(4 sinus of (x) plus nine) divide by (one plus ctg squared x)
(4sin(x)+9)/(1+ctg2x)
4sinx+9/1+ctg2x
(4sin(x)+9)/(1+ctg²x)
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4sinx+9/1+ctg^2x
(4sin(x)+9) divide by (1+ctg^2x)
Similar expressions
(4sin(x)-9)/(1+ctg^2x)
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(4sinx+9)/(1+ctg^2x)

The derivative of the function/ (4sin(x)+9)/(1+ctg^2x)

Derivative of (4sin(x)+9)/(1+ctg^2x)

The solution

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4*sin(x) + 9
------------
       2    
1 + cot (x)

$$\frac{4 \sin{\left(x \right)} + 9}{\cot^{2}{\left(x \right)} + 1}$$

(4*sin(x) + 9)/(1 + cot(x)^2)

Detail solution

Apply the quotient rule, which is:

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

$f{\left(x \right)} = 4 \sin{\left(x \right)} + 9$ and $g{\left(x \right)} = \cot^{2}{\left(x \right)} + 1$ .

To find $\frac{d}{d x} f{\left(x \right)}$ :
1. Differentiate $4 \sin{\left(x \right)} + 9$ term by term:
  1. The derivative of the constant $9$ is zero.
  2. The derivative of a constant times a function is the constant times the derivative of the function.
    1. The derivative of sine is cosine:
      
      $\frac{d}{d x} \sin{\left(x \right)} = \cos{\left(x \right)}$
    So, the result is: $4 \cos{\left(x \right)}$
  The result is: $4 \cos{\left(x \right)}$
To find $\frac{d}{d x} g{\left(x \right)}$ :
1. Differentiate $\cot^{2}{\left(x \right)} + 1$ term by term:
  1. The derivative of the constant $1$ is zero.
  2. Let $u = \cot{\left(x \right)}$ .
  3. Apply the power rule: $u^{2}$ goes to $2 u$
  4. Then, apply the chain rule. Multiply by $\frac{d}{d x} \cot{\left(x \right)}$ :
    1. There are multiple ways to do this derivative.
      Method #1
      
      Rewrite the function to be differentiated:
      
      $\cot{\left(x \right)} = \frac{1}{\tan{\left(x \right)}}$
      
      Let $u = \tan{\left(x \right)}$ .
      
      Apply the power rule: $\frac{1}{u}$ goes to $- \frac{1}{u^{2}}$
      
      Then, apply the chain rule. Multiply by $\frac{d}{d x} \tan{\left(x \right)}$ :
      
      Rewrite the function to be differentiated:
      
      $\tan{\left(x \right)} = \frac{\sin{\left(x \right)}}{\cos{\left(x \right)}}$
      
      Apply the quotient rule, which is:
      
      $\frac{d}{d x} \frac{f{\left(x \right)}}{g{\left(x \right)}} = \frac{- f{\left(x \right)} \frac{d}{d x} g{\left(x \right)} + g{\left(x \right)} \frac{d}{d x} f{\left(x \right)}}{g^{2}{\left(x \right)}}$
      
      $f{\left(x \right)} = \sin{\left(x \right)}$ and $g{\left(x \right)} = \cos{\left(x \right)}$ .
      
      To find $\frac{d}{d x} f{\left(x \right)}$ :
      
      The derivative of sine is cosine:
      
      $\frac{d}{d x} \sin{\left(x \right)} = \cos{\left(x \right)}$
      
      To find $\frac{d}{d x} g{\left(x \right)}$ :
      
      The derivative of cosine is negative sine:
      
      $\frac{d}{d x} \cos{\left(x \right)} = - \sin{\left(x \right)}$
      
      Now plug in to the quotient rule:
      
      $\frac{\sin^{2}{\left(x \right)} + \cos^{2}{\left(x \right)}}{\cos^{2}{\left(x \right)}}$
      
      The result of the chain rule is:
      
      $- \frac{\sin^{2}{\left(x \right)} + \cos^{2}{\left(x \right)}}{\cos^{2}{\left(x \right)} \tan^{2}{\left(x \right)}}$
      Method #2
      
      Rewrite the function to be differentiated:
      
      $\cot{\left(x \right)} = \frac{\cos{\left(x \right)}}{\sin{\left(x \right)}}$
      
      Apply the quotient rule, which is:
      
      $\frac{d}{d x} \frac{f{\left(x \right)}}{g{\left(x \right)}} = \frac{- f{\left(x \right)} \frac{d}{d x} g{\left(x \right)} + g{\left(x \right)} \frac{d}{d x} f{\left(x \right)}}{g^{2}{\left(x \right)}}$
      
      $f{\left(x \right)} = \cos{\left(x \right)}$ and $g{\left(x \right)} = \sin{\left(x \right)}$ .
      
      To find $\frac{d}{d x} f{\left(x \right)}$ :
      
      The derivative of cosine is negative sine:
      
      $\frac{d}{d x} \cos{\left(x \right)} = - \sin{\left(x \right)}$
      
      To find $\frac{d}{d x} g{\left(x \right)}$ :
      
      The derivative of sine is cosine:
      
      $\frac{d}{d x} \sin{\left(x \right)} = \cos{\left(x \right)}$
      
      Now plug in to the quotient rule:
      
      $\frac{- \sin^{2}{\left(x \right)} - \cos^{2}{\left(x \right)}}{\sin^{2}{\left(x \right)}}$
    The result of the chain rule is:
    
    $- \frac{2 \left(\sin^{2}{\left(x \right)} + \cos^{2}{\left(x \right)}\right) \cot{\left(x \right)}}{\cos^{2}{\left(x \right)} \tan^{2}{\left(x \right)}}$
  The result is: $- \frac{2 \left(\sin^{2}{\left(x \right)} + \cos^{2}{\left(x \right)}\right) \cot{\left(x \right)}}{\cos^{2}{\left(x \right)} \tan^{2}{\left(x \right)}}$
Now plug in to the quotient rule:

$\frac{\frac{2 \left(4 \sin{\left(x \right)} + 9\right) \left(\sin^{2}{\left(x \right)} + \cos^{2}{\left(x \right)}\right) \cot{\left(x \right)}}{\cos^{2}{\left(x \right)} \tan^{2}{\left(x \right)}} + 4 \left(\cot^{2}{\left(x \right)} + 1\right) \cos{\left(x \right)}}{\left(\cot^{2}{\left(x \right)} + 1\right)^{2}}$
Now simplify:

$6 \left(2 \sin{\left(x \right)} + 3\right) \sin{\left(x \right)} \cos{\left(x \right)}$

The answer is:

$6 \left(2 \sin{\left(x \right)} + 3\right) \sin{\left(x \right)} \cos{\left(x \right)}$

The graph

Plot the graph f(x)

Plot the graph f'(x)

The first derivative [src]

              /          2   \                      
  4*cos(x)    \-2 - 2*cot (x)/*(4*sin(x) + 9)*cot(x)
----------- - --------------------------------------
       2                               2            
1 + cot (x)               /       2   \             
                          \1 + cot (x)/

$$- \frac{\left(4 \sin{\left(x \right)} + 9\right) \left(- 2 \cot^{2}{\left(x \right)} - 2\right) \cot{\left(x \right)}}{\left(\cot^{2}{\left(x \right)} + 1\right)^{2}} + \frac{4 \cos{\left(x \right)}}{\cot^{2}{\left(x \right)} + 1}$$

Simplify

The second derivative [src]

  /  /          2    \                                               \
  |  |     2*cot (x) |                    2*sin(x)    8*cos(x)*cot(x)|
2*|- |1 - -----------|*(9 + 4*sin(x)) - ----------- + ---------------|
  |  |           2   |                         2               2     |
  \  \    1 + cot (x)/                  1 + cot (x)     1 + cot (x)  /

$$2 \left(- \left(1 - \frac{2 \cot^{2}{\left(x \right)}}{\cot^{2}{\left(x \right)} + 1}\right) \left(4 \sin{\left(x \right)} + 9\right) - \frac{2 \sin{\left(x \right)}}{\cot^{2}{\left(x \right)} + 1} + \frac{8 \cos{\left(x \right)} \cot{\left(x \right)}}{\cot^{2}{\left(x \right)} + 1}\right)$$

Simplify

The third derivative [src]

   /                /          2    \            /         2     \                                        \
   |   cos(x)       |     2*cot (x) |            |      cot (x)  |                         6*cot(x)*sin(x)|
-4*|----------- + 6*|1 - -----------|*cos(x) + 2*|1 - -----------|*(9 + 4*sin(x))*cot(x) + ---------------|
   |       2        |           2   |            |           2   |                                  2     |
   \1 + cot (x)     \    1 + cot (x)/            \    1 + cot (x)/                           1 + cot (x)  /

$$- 4 \left(6 \left(1 - \frac{2 \cot^{2}{\left(x \right)}}{\cot^{2}{\left(x \right)} + 1}\right) \cos{\left(x \right)} + 2 \left(1 - \frac{\cot^{2}{\left(x \right)}}{\cot^{2}{\left(x \right)} + 1}\right) \left(4 \sin{\left(x \right)} + 9\right) \cot{\left(x \right)} + \frac{6 \sin{\left(x \right)} \cot{\left(x \right)}}{\cot^{2}{\left(x \right)} + 1} + \frac{\cos{\left(x \right)}}{\cot^{2}{\left(x \right)} + 1}\right)$$

Simplify

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Identical expressions

Similar expressions

Derivative of (4sin(x)+9)/(1+ctg^2x)

The graph:

Piecewise:

The solution

Method #1

Method #2