Derivative of (cos^2)x

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   2     
cos (x)*x

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

cos(x)^2*x

Detail solution

Apply the product rule:

$\frac{d}{d x} f{\left(x \right)} g{\left(x \right)} = f{\left(x \right)} \frac{d}{d x} g{\left(x \right)} + g{\left(x \right)} \frac{d}{d x} f{\left(x \right)}$

$f{\left(x \right)} = \cos^{2}{\left(x \right)}$ ; to find $\frac{d}{d x} f{\left(x \right)}$ :
1. Let $u = \cos{\left(x \right)}$ .
2. Apply the power rule: $u^{2}$ goes to $2 u$
3. Then, apply the chain rule. Multiply by $\frac{d}{d x} \cos{\left(x \right)}$ :
  1. The derivative of cosine is negative sine:
    
    $\frac{d}{d x} \cos{\left(x \right)} = - \sin{\left(x \right)}$
  The result of the chain rule is:
  
  $- 2 \sin{\left(x \right)} \cos{\left(x \right)}$
$g{\left(x \right)} = x$ ; to find $\frac{d}{d x} g{\left(x \right)}$ :
1. Apply the power rule: $x$ goes to $1$
The result is: $- 2 x \sin{\left(x \right)} \cos{\left(x \right)} + \cos^{2}{\left(x \right)}$
Now simplify:

$- x \sin{\left(2 x \right)} + \frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}$

The answer is:

$- x \sin{\left(2 x \right)} + \frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}$

The graph

Plot the graph f(x)

Plot the graph f'(x)

The first derivative [src]

   2                       
cos (x) - 2*x*cos(x)*sin(x)

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

Simplify

The second derivative [src]

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

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

Simplify

The third derivative [src]

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

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

Simplify

The graph