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5sinx-4tanx
The derivative of the function/ 5sinx-4tanx

Derivative of 5sinx-4tanx

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You have entered [src] 
5*sin(x) - 4*tan(x)
5sin⁡(x)−4tan⁡(x)5 \sin{\left(x \right)} - 4 \tan{\left(x \right)} 
5*sin(x) - 4*tan(x)
Detail solution

  1. Differentiate 5sin⁡(x)−4tan⁡(x)5 \sin{\left(x \right)} - 4 \tan{\left(x \right)} term by term:

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

      1. The derivative of sine is cosine:

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

      So, the result is: 5cos⁡(x)5 \cos{\left(x \right)}

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

      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:

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

      So, the result is: −4(sin⁡2(x)+cos⁡2(x))cos⁡2(x)- \frac{4 \left(\sin^{2}{\left(x \right)} + \cos^{2}{\left(x \right)}\right)}{\cos^{2}{\left(x \right)}}

    The result is: −4(sin⁡2(x)+cos⁡2(x))cos⁡2(x)+5cos⁡(x)- \frac{4 \left(\sin^{2}{\left(x \right)} + \cos^{2}{\left(x \right)}\right)}{\cos^{2}{\left(x \right)}} + 5 \cos{\left(x \right)}

  2. Now simplify:

    5cos⁡(x)−4cos⁡2(x)5 \cos{\left(x \right)} - \frac{4}{\cos^{2}{\left(x \right)}}

The answer is:

5cos⁡(x)−4cos⁡2(x)5 \cos{\left(x \right)} - \frac{4}{\cos^{2}{\left(x \right)}}

The graph
02468-8-6-4-2-1010-50005000
Plot the graph f(x)
Plot the graph f'(x)
The first derivative [src] 
          2              
-4 - 4*tan (x) + 5*cos(x)
5cos⁡(x)−4tan⁡2(x)−45 \cos{\left(x \right)} - 4 \tan^{2}{\left(x \right)} - 4
Simplify
The second derivative [src] 
 /             /       2   \       \
-\5*sin(x) + 8*\1 + tan (x)/*tan(x)/
−(8(tan⁡2(x)+1)tan⁡(x)+5sin⁡(x))- (8 \left(\tan^{2}{\left(x \right)} + 1\right) \tan{\left(x \right)} + 5 \sin{\left(x \right)})
Simplify
The third derivative [src] 
 /                          2                           \
 |             /       2   \          2    /       2   \|
-\5*cos(x) + 8*\1 + tan (x)/  + 16*tan (x)*\1 + tan (x)//
−(8(tan⁡2(x)+1)2+16(tan⁡2(x)+1)tan⁡2(x)+5cos⁡(x))- (8 \left(\tan^{2}{\left(x \right)} + 1\right)^{2} + 16 \left(\tan^{2}{\left(x \right)} + 1\right) \tan^{2}{\left(x \right)} + 5 \cos{\left(x \right)})
Simplify
The graph
Derivative of 5sinx-4tanx
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