Derivative of (2cos(2t))/(sin(2t))

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2*cos(2*t)
----------
 sin(2*t)

$$\frac{2 \cos{\left(2 t \right)}}{\sin{\left(2 t \right)}}$$

(2*cos(2*t))/sin(2*t)

Detail solution

Apply the quotient rule, which is:

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

$f{\left(t \right)} = 2 \cos{\left(2 t \right)}$ and $g{\left(t \right)} = \sin{\left(2 t \right)}$ .

To find $\frac{d}{d t} f{\left(t \right)}$ :
1. The derivative of a constant times a function is the constant times the derivative of the function.
  1. Let $u = 2 t$ .
  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 $\frac{d}{d t} 2 t$ :
    1. The derivative of a constant times a function is the constant times the derivative of the function.
      1. Apply the power rule: $t$ goes to $1$
      So, the result is: $2$
    The result of the chain rule is:
    
    $- 2 \sin{\left(2 t \right)}$
  So, the result is: $- 4 \sin{\left(2 t \right)}$
To find $\frac{d}{d t} g{\left(t \right)}$ :
1. Let $u = 2 t$ .
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 $\frac{d}{d t} 2 t$ :
  1. The derivative of a constant times a function is the constant times the derivative of the function.
    1. Apply the power rule: $t$ goes to $1$
    So, the result is: $2$
  The result of the chain rule is:
  
  $2 \cos{\left(2 t \right)}$
Now plug in to the quotient rule:

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

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

The answer is:

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

The graph

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The first derivative [src]

          2     
     4*cos (2*t)
-4 - -----------
         2      
      sin (2*t)

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

Simplify

The second derivative [src]

  /         2     \         
  |    2*cos (2*t)|         
8*|2 + -----------|*cos(2*t)
  |        2      |         
  \     sin (2*t) /         
----------------------------
          sin(2*t)

$$\frac{8 \left(2 + \frac{2 \cos^{2}{\left(2 t \right)}}{\sin^{2}{\left(2 t \right)}}\right) \cos{\left(2 t \right)}}{\sin{\left(2 t \right)}}$$

Simplify

The third derivative [src]

    /                            /         2     \\
    |                     2      |    6*cos (2*t)||
    |                  cos (2*t)*|5 + -----------||
    |         2                  |        2      ||
    |    3*cos (2*t)             \     sin (2*t) /|
-16*|2 + ----------- + ---------------------------|
    |        2                     2              |
    \     sin (2*t)             sin (2*t)         /

$$- 16 \left(\frac{\left(5 + \frac{6 \cos^{2}{\left(2 t \right)}}{\sin^{2}{\left(2 t \right)}}\right) \cos^{2}{\left(2 t \right)}}{\sin^{2}{\left(2 t \right)}} + 2 + \frac{3 \cos^{2}{\left(2 t \right)}}{\sin^{2}{\left(2 t \right)}}\right)$$

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Derivative of (2cos(2t))/(sin(2t))

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The solution