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e^(4cos(x-1))

Derivative of e^(4cos(x-1))

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

The graph:

from to

Piecewise:

The solution

You have entered [src]
 4*cos(x - 1)
e            
$$e^{4 \cos{\left(x - 1 \right)}}$$
d / 4*cos(x - 1)\
--\e            /
dx               
$$\frac{d}{d x} e^{4 \cos{\left(x - 1 \right)}}$$
Detail solution
  1. Let .

  2. The derivative of is itself.

  3. Then, apply the chain rule. Multiply by :

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

      1. Let .

      2. The derivative of cosine is negative sine:

      3. Then, apply the chain rule. Multiply by :

        1. Differentiate term by term:

          1. Apply the power rule: goes to

          2. The derivative of the constant is zero.

          The result is:

        The result of the chain rule is:

      So, the result is:

    The result of the chain rule is:

  4. Now simplify:


The answer is:

The graph
The first derivative [src]
    4*cos(x - 1)           
-4*e            *sin(x - 1)
$$- 4 e^{4 \cos{\left(x - 1 \right)}} \sin{\left(x - 1 \right)}$$
The second derivative [src]
  /                    2        \  4*cos(-1 + x)
4*\-cos(-1 + x) + 4*sin (-1 + x)/*e             
$$4 \cdot \left(4 \sin^{2}{\left(x - 1 \right)} - \cos{\left(x - 1 \right)}\right) e^{4 \cos{\left(x - 1 \right)}}$$
The third derivative [src]
  /          2                         \  4*cos(-1 + x)            
4*\1 - 16*sin (-1 + x) + 12*cos(-1 + x)/*e             *sin(-1 + x)
$$4 \left(- 16 \sin^{2}{\left(x - 1 \right)} + 12 \cos{\left(x - 1 \right)} + 1\right) e^{4 \cos{\left(x - 1 \right)}} \sin{\left(x - 1 \right)}$$
The graph
Derivative of e^(4cos(x-1))