Mister Exam

Derivative of y=tg2x^5

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

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Piecewise:

The solution

You have entered [src]
   5     
tan (2*x)
tan5(2x)\tan^{5}{\left(2 x \right)}
d /   5     \
--\tan (2*x)/
dx           
ddxtan5(2x)\frac{d}{d x} \tan^{5}{\left(2 x \right)}
Detail solution
  1. Let u=tan(2x)u = \tan{\left(2 x \right)}.

  2. Apply the power rule: u5u^{5} goes to 5u45 u^{4}

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

    1. Rewrite the function to be differentiated:

      tan(2x)=sin(2x)cos(2x)\tan{\left(2 x \right)} = \frac{\sin{\left(2 x \right)}}{\cos{\left(2 x \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)f{\left(x \right)} = \sin{\left(2 x \right)} and g(x)=cos(2x)g{\left(x \right)} = \cos{\left(2 x \right)}.

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

      1. Let u=2xu = 2 x.

      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 ddx2x\frac{d}{d x} 2 x:

        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

        The result of the chain rule is:

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

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

      1. Let u=2xu = 2 x.

      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 ddx2x\frac{d}{d x} 2 x:

        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

        The result of the chain rule is:

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

      Now plug in to the quotient rule:

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

    The result of the chain rule is:

    5(2sin2(2x)+2cos2(2x))tan4(2x)cos2(2x)\frac{5 \cdot \left(2 \sin^{2}{\left(2 x \right)} + 2 \cos^{2}{\left(2 x \right)}\right) \tan^{4}{\left(2 x \right)}}{\cos^{2}{\left(2 x \right)}}

  4. Now simplify:

    10tan4(2x)cos2(2x)\frac{10 \tan^{4}{\left(2 x \right)}}{\cos^{2}{\left(2 x \right)}}


The answer is:

10tan4(2x)cos2(2x)\frac{10 \tan^{4}{\left(2 x \right)}}{\cos^{2}{\left(2 x \right)}}

The graph
02468-8-6-4-2-1010-5000000000050000000000
The first derivative [src]
   4      /           2     \
tan (2*x)*\10 + 10*tan (2*x)/
(10tan2(2x)+10)tan4(2x)\left(10 \tan^{2}{\left(2 x \right)} + 10\right) \tan^{4}{\left(2 x \right)}
The second derivative [src]
      3      /       2     \ /         2     \
40*tan (2*x)*\1 + tan (2*x)/*\2 + 3*tan (2*x)/
40(tan2(2x)+1)(3tan2(2x)+2)tan3(2x)40 \left(\tan^{2}{\left(2 x \right)} + 1\right) \left(3 \tan^{2}{\left(2 x \right)} + 2\right) \tan^{3}{\left(2 x \right)}
The third derivative [src]
                             /                               2                               \
      2      /       2     \ |     4          /       2     \          2      /       2     \|
80*tan (2*x)*\1 + tan (2*x)/*\2*tan (2*x) + 6*\1 + tan (2*x)/  + 13*tan (2*x)*\1 + tan (2*x)//
80(tan2(2x)+1)(6(tan2(2x)+1)2+13(tan2(2x)+1)tan2(2x)+2tan4(2x))tan2(2x)80 \left(\tan^{2}{\left(2 x \right)} + 1\right) \left(6 \left(\tan^{2}{\left(2 x \right)} + 1\right)^{2} + 13 \left(\tan^{2}{\left(2 x \right)} + 1\right) \tan^{2}{\left(2 x \right)} + 2 \tan^{4}{\left(2 x \right)}\right) \tan^{2}{\left(2 x \right)}
The graph
Derivative of y=tg2x^5