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Derivative of cos(x^2-4*x)

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   / 2      \
cos\x  - 4*x/

$$\cos{\left(x^{2} - 4 x \right)}$$

cos(x^2 - 4*x)

Detail solution

Let $u = x^{2} - 4 x$ .
The derivative of cosine is negative sine:

$\frac{d}{d u} \cos{\left(u \right)} = - \sin{\left(u \right)}$
Then, apply the chain rule. Multiply by $\frac{d}{d x} \left(x^{2} - 4 x\right)$ :
1. Differentiate $x^{2} - 4 x$ term by term:
  1. Apply the power rule: $x^{2}$ goes to $2 x$
  2. The derivative of a constant times a function is the constant times the derivative of the function.
    1. Apply the power rule: $x$ goes to $1$
    So, the result is: $-4$
  The result is: $2 x - 4$
The result of the chain rule is:

$- \left(2 x - 4\right) \sin{\left(x^{2} - 4 x \right)}$
Now simplify:

$\left(4 - 2 x\right) \sin{\left(x \left(x - 4\right) \right)}$

The answer is:

$\left(4 - 2 x\right) \sin{\left(x \left(x - 4\right) \right)}$

The graph

Plot the graph f(x)

Plot the graph f'(x)

The first derivative [src]

               / 2      \
-(-4 + 2*x)*sin\x  - 4*x/

$$- \left(2 x - 4\right) \sin{\left(x^{2} - 4 x \right)}$$

Simplify

The second derivative [src]

   /          2                                  \
-2*\2*(-2 + x) *cos(x*(-4 + x)) + sin(x*(-4 + x))/

$$- 2 \left(2 \left(x - 2\right)^{2} \cos{\left(x \left(x - 4\right) \right)} + \sin{\left(x \left(x - 4\right) \right)}\right)$$

Simplify

The third derivative [src]

           /                               2                \
4*(-2 + x)*\-3*cos(x*(-4 + x)) + 2*(-2 + x) *sin(x*(-4 + x))/

$$4 \left(x - 2\right) \left(2 \left(x - 2\right)^{2} \sin{\left(x \left(x - 4\right) \right)} - 3 \cos{\left(x \left(x - 4\right) \right)}\right)$$

Simplify

The graph