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The derivative of the function/ 2*t*sin(4*x)+1/x

Derivative of 2*t*sin(4*x)+1/x

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

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The solution

You have entered [src] 
               1
2*t*sin(4*x) + -
               x
2tsin⁡(4x)+1x2 t \sin{\left(4 x \right)} + \frac{1}{x} 
(2*t)*sin(4*x) + 1/x
Detail solution

  1. Differentiate 2tsin⁡(4x)+1x2 t \sin{\left(4 x \right)} + \frac{1}{x} term by term:

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

      1. Let u=4xu = 4 x.

      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 ddx4x\frac{d}{d x} 4 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: 44

        The result of the chain rule is:

        4cos⁡(4x)4 \cos{\left(4 x \right)}

      So, the result is: 8tcos⁡(4x)8 t \cos{\left(4 x \right)}

    2. Apply the power rule: 1x\frac{1}{x} goes to −1x2- \frac{1}{x^{2}}

    The result is: 8tcos⁡(4x)−1x28 t \cos{\left(4 x \right)} - \frac{1}{x^{2}}

The answer is:

8tcos⁡(4x)−1x28 t \cos{\left(4 x \right)} - \frac{1}{x^{2}}

The first derivative [src] 
  1                
- -- + 8*t*cos(4*x)
   2               
  x
8tcos⁡(4x)−1x28 t \cos{\left(4 x \right)} - \frac{1}{x^{2}}
Simplify
The second derivative [src] 
  /1                 \
2*|-- - 16*t*sin(4*x)|
  | 3                |
  \x                 /
2(−16tsin⁡(4x)+1x3)2 \left(- 16 t \sin{\left(4 x \right)} + \frac{1}{x^{3}}\right)
Simplify
The third derivative [src] 
   /3                 \
-2*|-- + 64*t*cos(4*x)|
   | 4                |
   \x                 /
−2(64tcos⁡(4x)+3x4)- 2 \left(64 t \cos{\left(4 x \right)} + \frac{3}{x^{4}}\right)
Simplify
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