Derivative of cos(z(z+1))

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cos(z*(z + 1))

$$\cos{\left(z \left(z + 1\right) \right)}$$

cos(z*(z + 1))

Detail solution

Let $u = z \left(z + 1\right)$ .
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 $\frac{d}{d z} z \left(z + 1\right)$ :
1. Apply the product rule:
  
  $\frac{d}{d z} f{\left(z \right)} g{\left(z \right)} = f{\left(z \right)} \frac{d}{d z} g{\left(z \right)} + g{\left(z \right)} \frac{d}{d z} f{\left(z \right)}$
  
  $f{\left(z \right)} = z$ ; to find $\frac{d}{d z} f{\left(z \right)}$ :
  1. Apply the power rule: $z$ goes to $1$
  $g{\left(z \right)} = z + 1$ ; to find $\frac{d}{d z} g{\left(z \right)}$ :
  1. Differentiate $z + 1$ term by term:
    1. Apply the power rule: $z$ goes to $1$
    2. The derivative of the constant $1$ is zero.
    The result is: $1$
  The result is: $2 z + 1$
The result of the chain rule is:

$- \left(2 z + 1\right) \sin{\left(z \left(z + 1\right) \right)}$
Now simplify:

$- \left(2 z + 1\right) \sin{\left(z \left(z + 1\right) \right)}$

The answer is:

$- \left(2 z + 1\right) \sin{\left(z \left(z + 1\right) \right)}$

The graph

Plot the graph f(x)

Plot the graph f'(x)

The first derivative [src]

-(1 + 2*z)*sin(z*(z + 1))

$$- \left(2 z + 1\right) \sin{\left(z \left(z + 1\right) \right)}$$

Simplify

The second derivative [src]

 /                            2               \
-\2*sin(z*(1 + z)) + (1 + 2*z) *cos(z*(1 + z))/

$$- (\left(2 z + 1\right)^{2} \cos{\left(z \left(z + 1\right) \right)} + 2 \sin{\left(z \left(z + 1\right) \right)})$$

Simplify

The third derivative [src]

          /                             2               \
(1 + 2*z)*\-6*cos(z*(1 + z)) + (1 + 2*z) *sin(z*(1 + z))/

$$\left(2 z + 1\right) \left(\left(2 z + 1\right)^{2} \sin{\left(z \left(z + 1\right) \right)} - 6 \cos{\left(z \left(z + 1\right) \right)}\right)$$

Simplify