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e^(7*x)*cot(x)

Derivative of e^(7*x)*cot(x)

Function f() - derivative -N order at the point
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You have entered [src]
 7*x       
E   *cot(x)
e7xcot(x)e^{7 x} \cot{\left(x \right)}
E^(7*x)*cot(x)
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)=e7xf{\left(x \right)} = e^{7 x}; to find ddxf(x)\frac{d}{d x} f{\left(x \right)}:

    1. Let u=7xu = 7 x.

    2. The derivative of eue^{u} is itself.

    3. Then, apply the chain rule. Multiply by ddx7x\frac{d}{d x} 7 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: 77

      The result of the chain rule is:

      7e7x7 e^{7 x}

    g(x)=cot(x)g{\left(x \right)} = \cot{\left(x \right)}; to find ddxg(x)\frac{d}{d x} g{\left(x \right)}:

    1. There are multiple ways to do this derivative.

      Method #1

      1. Rewrite the function to be differentiated:

        cot(x)=1tan(x)\cot{\left(x \right)} = \frac{1}{\tan{\left(x \right)}}

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

      3. Apply the power rule: 1u\frac{1}{u} goes to 1u2- \frac{1}{u^{2}}

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

        1. Rewrite the function to be differentiated:

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

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

          1. The derivative of sine is cosine:

            ddxsin(x)=cos(x)\frac{d}{d x} \sin{\left(x \right)} = \cos{\left(x \right)}

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

          1. The derivative of cosine is negative sine:

            ddxcos(x)=sin(x)\frac{d}{d x} \cos{\left(x \right)} = - \sin{\left(x \right)}

          Now plug in to the quotient rule:

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

        The result of the chain rule is:

        sin2(x)+cos2(x)cos2(x)tan2(x)- \frac{\sin^{2}{\left(x \right)} + \cos^{2}{\left(x \right)}}{\cos^{2}{\left(x \right)} \tan^{2}{\left(x \right)}}

      Method #2

      1. Rewrite the function to be differentiated:

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

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

        1. The derivative of cosine is negative sine:

          ddxcos(x)=sin(x)\frac{d}{d x} \cos{\left(x \right)} = - \sin{\left(x \right)}

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

        1. The derivative of sine is cosine:

          ddxsin(x)=cos(x)\frac{d}{d x} \sin{\left(x \right)} = \cos{\left(x \right)}

        Now plug in to the quotient rule:

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

    The result is: (sin2(x)+cos2(x))e7xcos2(x)tan2(x)+7e7xcot(x)- \frac{\left(\sin^{2}{\left(x \right)} + \cos^{2}{\left(x \right)}\right) e^{7 x}}{\cos^{2}{\left(x \right)} \tan^{2}{\left(x \right)}} + 7 e^{7 x} \cot{\left(x \right)}

  2. Now simplify:

    (7sin(2x)2)e7x1cos(2x)\frac{\left(7 \sin{\left(2 x \right)} - 2\right) e^{7 x}}{1 - \cos{\left(2 x \right)}}


The answer is:

(7sin(2x)2)e7x1cos(2x)\frac{\left(7 \sin{\left(2 x \right)} - 2\right) e^{7 x}}{1 - \cos{\left(2 x \right)}}

The graph
02468-8-6-4-2-1010-5e315e31
The first derivative [src]
/        2   \  7*x             7*x
\-1 - cot (x)/*e    + 7*cot(x)*e   
(cot2(x)1)e7x+7e7xcot(x)\left(- \cot^{2}{\left(x \right)} - 1\right) e^{7 x} + 7 e^{7 x} \cot{\left(x \right)}
The second derivative [src]
/            2                    /       2   \       \  7*x
\-14 - 14*cot (x) + 49*cot(x) + 2*\1 + cot (x)/*cot(x)/*e   
(2(cot2(x)+1)cot(x)14cot2(x)+49cot(x)14)e7x\left(2 \left(\cot^{2}{\left(x \right)} + 1\right) \cot{\left(x \right)} - 14 \cot^{2}{\left(x \right)} + 49 \cot{\left(x \right)} - 14\right) e^{7 x}
The third derivative [src]
/              2                     /       2   \ /         2   \      /       2   \       \  7*x
\-147 - 147*cot (x) + 343*cot(x) - 2*\1 + cot (x)/*\1 + 3*cot (x)/ + 42*\1 + cot (x)/*cot(x)/*e   
(2(cot2(x)+1)(3cot2(x)+1)+42(cot2(x)+1)cot(x)147cot2(x)+343cot(x)147)e7x\left(- 2 \left(\cot^{2}{\left(x \right)} + 1\right) \left(3 \cot^{2}{\left(x \right)} + 1\right) + 42 \left(\cot^{2}{\left(x \right)} + 1\right) \cot{\left(x \right)} - 147 \cot^{2}{\left(x \right)} + 343 \cot{\left(x \right)} - 147\right) e^{7 x}
The graph
Derivative of e^(7*x)*cot(x)