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Derivative of e
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Derivative of 1/(1-x)
Identical expressions
y=ln(tgx^ two)
y equally ln(tgx squared )
y equally ln(tgx to the power of two)
y=ln(tgx2)
y=lntgx2
y=ln(tgx²)
y=ln(tgx to the power of 2)
y=lntgx^2

The derivative of the function/ y=ln(tgx^2)

Derivative of y=ln(tgx^2)

The solution

You have entered [src]

   /   2   \
log\tan (x)/

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

log(tan(x)^2)

Detail solution

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

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

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

The answer is:

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

The graph

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Plot the graph f'(x)

The first derivative [src]

         2   
2 + 2*tan (x)
-------------
    tan(x)

$$\frac{2 \tan^{2}{\left(x \right)} + 2}{\tan{\left(x \right)}}$$

Simplify

The second derivative [src]

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

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

Simplify

The third derivative [src]

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

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

Simplify

The graph

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Derivative of y=ln(tgx^2)

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The solution