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Derivative of x^2*(tan(4x+1))

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

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 2             
x *tan(4*x + 1)
x2tan(4x+1)x^{2} \tan{\left(4 x + 1 \right)}
x^2*tan(4*x + 1)
Detail solution
  1. Apply the product rule:

    ddxf(x)g(x)=f(x)ddxg(x)+g(x)ddxf(x)\frac{d}{d x} f{\left(x \right)} g{\left(x \right)} = f{\left(x \right)} \frac{d}{d x} g{\left(x \right)} + g{\left(x \right)} \frac{d}{d x} f{\left(x \right)}

    f(x)=x2f{\left(x \right)} = x^{2}; to find ddxf(x)\frac{d}{d x} f{\left(x \right)}:

    1. Apply the power rule: x2x^{2} goes to 2x2 x

    g(x)=tan(4x+1)g{\left(x \right)} = \tan{\left(4 x + 1 \right)}; to find ddxg(x)\frac{d}{d x} g{\left(x \right)}:

    1. Rewrite the function to be differentiated:

      tan(4x+1)=sin(4x+1)cos(4x+1)\tan{\left(4 x + 1 \right)} = \frac{\sin{\left(4 x + 1 \right)}}{\cos{\left(4 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(4x+1)f{\left(x \right)} = \sin{\left(4 x + 1 \right)} and g(x)=cos(4x+1)g{\left(x \right)} = \cos{\left(4 x + 1 \right)}.

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

      1. Let u=4x+1u = 4 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(4x+1)\frac{d}{d x} \left(4 x + 1\right):

        1. Differentiate 4x+14 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: 44

          2. The derivative of the constant 11 is zero.

          The result is: 44

        The result of the chain rule is:

        4cos(4x+1)4 \cos{\left(4 x + 1 \right)}

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

      1. Let u=4x+1u = 4 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(4x+1)\frac{d}{d x} \left(4 x + 1\right):

        1. Differentiate 4x+14 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: 44

          2. The derivative of the constant 11 is zero.

          The result is: 44

        The result of the chain rule is:

        4sin(4x+1)- 4 \sin{\left(4 x + 1 \right)}

      Now plug in to the quotient rule:

      4sin2(4x+1)+4cos2(4x+1)cos2(4x+1)\frac{4 \sin^{2}{\left(4 x + 1 \right)} + 4 \cos^{2}{\left(4 x + 1 \right)}}{\cos^{2}{\left(4 x + 1 \right)}}

    The result is: x2(4sin2(4x+1)+4cos2(4x+1))cos2(4x+1)+2xtan(4x+1)\frac{x^{2} \left(4 \sin^{2}{\left(4 x + 1 \right)} + 4 \cos^{2}{\left(4 x + 1 \right)}\right)}{\cos^{2}{\left(4 x + 1 \right)}} + 2 x \tan{\left(4 x + 1 \right)}

  2. Now simplify:

    4x2cos2(4x+1)+2xtan(4x+1)\frac{4 x^{2}}{\cos^{2}{\left(4 x + 1 \right)}} + 2 x \tan{\left(4 x + 1 \right)}


The answer is:

4x2cos2(4x+1)+2xtan(4x+1)\frac{4 x^{2}}{\cos^{2}{\left(4 x + 1 \right)}} + 2 x \tan{\left(4 x + 1 \right)}

The graph
02468-8-6-4-2-1010-50000005000000
The first derivative [src]
 2 /         2         \                   
x *\4 + 4*tan (4*x + 1)/ + 2*x*tan(4*x + 1)
x2(4tan2(4x+1)+4)+2xtan(4x+1)x^{2} \left(4 \tan^{2}{\left(4 x + 1 \right)} + 4\right) + 2 x \tan{\left(4 x + 1 \right)}
The second derivative [src]
  /    /       2         \       2 /       2         \                            \
2*\8*x*\1 + tan (1 + 4*x)/ + 16*x *\1 + tan (1 + 4*x)/*tan(1 + 4*x) + tan(1 + 4*x)/
2(16x2(tan2(4x+1)+1)tan(4x+1)+8x(tan2(4x+1)+1)+tan(4x+1))2 \left(16 x^{2} \left(\tan^{2}{\left(4 x + 1 \right)} + 1\right) \tan{\left(4 x + 1 \right)} + 8 x \left(\tan^{2}{\left(4 x + 1 \right)} + 1\right) + \tan{\left(4 x + 1 \right)}\right)
The third derivative [src]
  /         2                2 /       2         \ /         2         \        /       2         \             \
8*\3 + 3*tan (1 + 4*x) + 16*x *\1 + tan (1 + 4*x)/*\1 + 3*tan (1 + 4*x)/ + 24*x*\1 + tan (1 + 4*x)/*tan(1 + 4*x)/
8(16x2(tan2(4x+1)+1)(3tan2(4x+1)+1)+24x(tan2(4x+1)+1)tan(4x+1)+3tan2(4x+1)+3)8 \left(16 x^{2} \left(\tan^{2}{\left(4 x + 1 \right)} + 1\right) \left(3 \tan^{2}{\left(4 x + 1 \right)} + 1\right) + 24 x \left(\tan^{2}{\left(4 x + 1 \right)} + 1\right) \tan{\left(4 x + 1 \right)} + 3 \tan^{2}{\left(4 x + 1 \right)} + 3\right)