Mister Exam

Derivative of (tg(x)-ctg(x))

Function f() - derivative -N order at the point
v

The graph:

from to

Piecewise:

The solution

You have entered [src]
tan(x) - cot(x)
tan(x)cot(x)\tan{\left(x \right)} - \cot{\left(x \right)}
d                  
--(tan(x) - cot(x))
dx                 
ddx(tan(x)cot(x))\frac{d}{d x} \left(\tan{\left(x \right)} - \cot{\left(x \right)}\right)
Detail solution
  1. Differentiate tan(x)cot(x)\tan{\left(x \right)} - \cot{\left(x \right)} term by term:

    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)}}

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

      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. ddxtan(x)=1cos2(x)\frac{d}{d x} \tan{\left(x \right)} = \frac{1}{\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)}}

      So, the result 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)}}

    The result is: sin2(x)+cos2(x)cos2(x)+sin2(x)+cos2(x)cos2(x)tan2(x)\frac{\sin^{2}{\left(x \right)} + \cos^{2}{\left(x \right)}}{\cos^{2}{\left(x \right)}} + \frac{\sin^{2}{\left(x \right)} + \cos^{2}{\left(x \right)}}{\cos^{2}{\left(x \right)} \tan^{2}{\left(x \right)}}

  2. Now simplify:

    81cos(4x)\frac{8}{1 - \cos{\left(4 x \right)}}


The answer is:

81cos(4x)\frac{8}{1 - \cos{\left(4 x \right)}}

The graph
02468-8-6-4-2-1010-1000010000
The first derivative [src]
       2         2   
2 + cot (x) + tan (x)
tan2(x)+cot2(x)+2\tan^{2}{\left(x \right)} + \cot^{2}{\left(x \right)} + 2
The second derivative [src]
  //       2   \          /       2   \       \
2*\\1 + tan (x)/*tan(x) - \1 + cot (x)/*cot(x)/
2((tan2(x)+1)tan(x)(cot2(x)+1)cot(x))2 \left(\left(\tan^{2}{\left(x \right)} + 1\right) \tan{\left(x \right)} - \left(\cot^{2}{\left(x \right)} + 1\right) \cot{\left(x \right)}\right)
The third derivative [src]
  /             2                2                                                    \
  |/       2   \    /       2   \         2    /       2   \        2    /       2   \|
2*\\1 + cot (x)/  + \1 + tan (x)/  + 2*cot (x)*\1 + cot (x)/ + 2*tan (x)*\1 + tan (x)//
2(2(tan2(x)+1)tan2(x)+2(cot2(x)+1)cot2(x)+(tan2(x)+1)2+(cot2(x)+1)2)2 \left(2 \left(\tan^{2}{\left(x \right)} + 1\right) \tan^{2}{\left(x \right)} + 2 \left(\cot^{2}{\left(x \right)} + 1\right) \cot^{2}{\left(x \right)} + \left(\tan^{2}{\left(x \right)} + 1\right)^{2} + \left(\cot^{2}{\left(x \right)} + 1\right)^{2}\right)
The graph
Derivative of (tg(x)-ctg(x))