Mister Exam

Derivative of lntg(2x+1)/4

Function f() - derivative -N order at the point
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The solution

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log(tan(2*x + 1))
-----------------
        4        
log(tan(2x+1))4\frac{\log{\left(\tan{\left(2 x + 1 \right)} \right)}}{4}
log(tan(2*x + 1))/4
Detail solution
  1. The derivative of a constant times a function is the constant times the derivative of the function.

    1. Let u=tan(2x+1)u = \tan{\left(2 x + 1 \right)}.

    2. The derivative of log(u)\log{\left(u \right)} is 1u\frac{1}{u}.

    3. Then, apply the chain rule. Multiply by ddxtan(2x+1)\frac{d}{d x} \tan{\left(2 x + 1 \right)}:

      1. Rewrite the function to be differentiated:

        tan(2x+1)=sin(2x+1)cos(2x+1)\tan{\left(2 x + 1 \right)} = \frac{\sin{\left(2 x + 1 \right)}}{\cos{\left(2 x + 1 \right)}}

      2. Apply the quotient rule, which is:

        ddxf(x)g(x)=f(x)ddxg(x)+g(x)ddxf(x)g2(x)\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(x)=sin(2x+1)f{\left(x \right)} = \sin{\left(2 x + 1 \right)} and g(x)=cos(2x+1)g{\left(x \right)} = \cos{\left(2 x + 1 \right)}.

        To find ddxf(x)\frac{d}{d x} f{\left(x \right)}:

        1. Let u=2x+1u = 2 x + 1.

        2. The derivative of sine is cosine:

          ddusin(u)=cos(u)\frac{d}{d u} \sin{\left(u \right)} = \cos{\left(u \right)}

        3. Then, apply the chain rule. Multiply by ddx(2x+1)\frac{d}{d x} \left(2 x + 1\right):

          1. Differentiate 2x+12 x + 1 term by term:

            1. The derivative of a constant times a function is the constant times the derivative of the function.

              1. Apply the power rule: xx goes to 11

              So, the result is: 22

            2. The derivative of the constant 11 is zero.

            The result is: 22

          The result of the chain rule is:

          2cos(2x+1)2 \cos{\left(2 x + 1 \right)}

        To find ddxg(x)\frac{d}{d x} g{\left(x \right)}:

        1. Let u=2x+1u = 2 x + 1.

        2. The derivative of cosine is negative sine:

          dducos(u)=sin(u)\frac{d}{d u} \cos{\left(u \right)} = - \sin{\left(u \right)}

        3. Then, apply the chain rule. Multiply by ddx(2x+1)\frac{d}{d x} \left(2 x + 1\right):

          1. Differentiate 2x+12 x + 1 term by term:

            1. The derivative of a constant times a function is the constant times the derivative of the function.

              1. Apply the power rule: xx goes to 11

              So, the result is: 22

            2. The derivative of the constant 11 is zero.

            The result is: 22

          The result of the chain rule is:

          2sin(2x+1)- 2 \sin{\left(2 x + 1 \right)}

        Now plug in to the quotient rule:

        2sin2(2x+1)+2cos2(2x+1)cos2(2x+1)\frac{2 \sin^{2}{\left(2 x + 1 \right)} + 2 \cos^{2}{\left(2 x + 1 \right)}}{\cos^{2}{\left(2 x + 1 \right)}}

      The result of the chain rule is:

      2sin2(2x+1)+2cos2(2x+1)cos2(2x+1)tan(2x+1)\frac{2 \sin^{2}{\left(2 x + 1 \right)} + 2 \cos^{2}{\left(2 x + 1 \right)}}{\cos^{2}{\left(2 x + 1 \right)} \tan{\left(2 x + 1 \right)}}

    So, the result is: 2sin2(2x+1)+2cos2(2x+1)4cos2(2x+1)tan(2x+1)\frac{2 \sin^{2}{\left(2 x + 1 \right)} + 2 \cos^{2}{\left(2 x + 1 \right)}}{4 \cos^{2}{\left(2 x + 1 \right)} \tan{\left(2 x + 1 \right)}}

  2. Now simplify:

    12cos2(2x+1)tan(2x+1)\frac{1}{2 \cos^{2}{\left(2 x + 1 \right)} \tan{\left(2 x + 1 \right)}}


The answer is:

12cos2(2x+1)tan(2x+1)\frac{1}{2 \cos^{2}{\left(2 x + 1 \right)} \tan{\left(2 x + 1 \right)}}

The graph
02468-8-6-4-2-1010-100100
The first derivative [src]
         2         
2 + 2*tan (2*x + 1)
-------------------
   4*tan(2*x + 1)  
2tan2(2x+1)+24tan(2x+1)\frac{2 \tan^{2}{\left(2 x + 1 \right)} + 2}{4 \tan{\left(2 x + 1 \right)}}
The second derivative [src]
                                         2
                      /       2         \ 
         2            \1 + tan (1 + 2*x)/ 
2 + 2*tan (1 + 2*x) - --------------------
                            2             
                         tan (1 + 2*x)    
(tan2(2x+1)+1)2tan2(2x+1)+2tan2(2x+1)+2- \frac{\left(\tan^{2}{\left(2 x + 1 \right)} + 1\right)^{2}}{\tan^{2}{\left(2 x + 1 \right)}} + 2 \tan^{2}{\left(2 x + 1 \right)} + 2
The third derivative [src]
                      /                                    2                        \
                      |                 /       2         \      /       2         \|
  /       2         \ |                 \1 + tan (1 + 2*x)/    2*\1 + tan (1 + 2*x)/|
4*\1 + tan (1 + 2*x)/*|2*tan(1 + 2*x) + -------------------- - ---------------------|
                      |                       3                     tan(1 + 2*x)    |
                      \                    tan (1 + 2*x)                            /
4(tan2(2x+1)+1)((tan2(2x+1)+1)2tan3(2x+1)2(tan2(2x+1)+1)tan(2x+1)+2tan(2x+1))4 \left(\tan^{2}{\left(2 x + 1 \right)} + 1\right) \left(\frac{\left(\tan^{2}{\left(2 x + 1 \right)} + 1\right)^{2}}{\tan^{3}{\left(2 x + 1 \right)}} - \frac{2 \left(\tan^{2}{\left(2 x + 1 \right)} + 1\right)}{\tan{\left(2 x + 1 \right)}} + 2 \tan{\left(2 x + 1 \right)}\right)
The graph
Derivative of lntg(2x+1)/4