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y=sinx/2cos^2x
The derivative of the function/ y=sinx/2cos^2x

Derivative of y=sinx/2cos^2x

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

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The solution

You have entered [src] 
sin(x)    2   
------*cos (x)
  2
sin⁡(x)2cos⁡2(x)\frac{\sin{\left(x \right)}}{2} \cos^{2}{\left(x \right)} 
(sin(x)/2)*cos(x)^2
Detail solution

  1. 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)cos⁡2(x)f{\left(x \right)} = \sin{\left(x \right)} \cos^{2}{\left(x \right)} and g(x)=2g{\left(x \right)} = 2.

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

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

      1. Let u=cos⁡(x)u = \cos{\left(x \right)}.

      2. Apply the power rule: u2u^{2} goes to 2u2 u

      3. Then, apply the chain rule. Multiply by ddxcos⁡(x)\frac{d}{d x} \cos{\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)}

        The result of the chain rule is:

        −2sin⁡(x)cos⁡(x)- 2 \sin{\left(x \right)} \cos{\left(x \right)}

      g(x)=sin⁡(x)g{\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)}

      The result is: −2sin⁡2(x)cos⁡(x)+cos⁡3(x)- 2 \sin^{2}{\left(x \right)} \cos{\left(x \right)} + \cos^{3}{\left(x \right)}

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

    1. The derivative of the constant 22 is zero.

    Now plug in to the quotient rule:

    −sin⁡2(x)cos⁡(x)+cos⁡3(x)2- \sin^{2}{\left(x \right)} \cos{\left(x \right)} + \frac{\cos^{3}{\left(x \right)}}{2}

  2. Now simplify:

    cos⁡(x)8+3cos⁡(3x)8\frac{\cos{\left(x \right)}}{8} + \frac{3 \cos{\left(3 x \right)}}{8}

The answer is:

cos⁡(x)8+3cos⁡(3x)8\frac{\cos{\left(x \right)}}{8} + \frac{3 \cos{\left(3 x \right)}}{8}

The graph
02468-8-6-4-2-10101-1
Plot the graph f(x)
Plot the graph f'(x)
The first derivative [src] 
   3                    
cos (x)      2          
------- - sin (x)*cos(x)
   2
−sin⁡2(x)cos⁡(x)+cos⁡3(x)2- \sin^{2}{\left(x \right)} \cos{\left(x \right)} + \frac{\cos^{3}{\left(x \right)}}{2}
Simplify
The second derivative [src] 
/               2   \       
|   2      7*cos (x)|       
|sin (x) - ---------|*sin(x)
\              2    /
(sin⁡2(x)−7cos⁡2(x)2)sin⁡(x)\left(\sin^{2}{\left(x \right)} - \frac{7 \cos^{2}{\left(x \right)}}{2}\right) \sin{\left(x \right)}
Simplify
The third derivative [src] 
/                  2   \       
|      2      7*cos (x)|       
|10*sin (x) - ---------|*cos(x)
\                 2    /
(10sin⁡2(x)−7cos⁡2(x)2)cos⁡(x)\left(10 \sin^{2}{\left(x \right)} - \frac{7 \cos^{2}{\left(x \right)}}{2}\right) \cos{\left(x \right)}
Simplify
The graph
Derivative of y=sinx/2cos^2x
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