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Identical expressions
x*ln(x+ one)-cos4x
x multiply by ln(x plus 1) minus co sinus of e of 4x
x multiply by ln(x plus one) minus co sinus of e of 4x
xln(x+1)-cos4x
xlnx+1-cos4x
Similar expressions
x*ln(x+1)+cos4x
x*ln(x-1)-cos4x

The derivative of the function/ x*ln(x+1)-cos4x

Derivative of x*ln(x+1)-cos4x

The solution

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x*log(x + 1) - cos(4*x)

$$x \log{\left(x + 1 \right)} - \cos{\left(4 x \right)}$$

x*log(x + 1) - cos(4*x)

Detail solution

Differentiate $x \log{\left(x + 1 \right)} - \cos{\left(4 x \right)}$ term by term:
1. Apply the product rule:
  
  $\frac{d}{d x} f{\left(x \right)} g{\left(x \right)} = f{\left(x \right)} \frac{d}{d x} g{\left(x \right)} + g{\left(x \right)} \frac{d}{d x} f{\left(x \right)}$
  
  $f{\left(x \right)} = x$ ; to find $\frac{d}{d x} f{\left(x \right)}$ :
  1. Apply the power rule: $x$ goes to $1$
  $g{\left(x \right)} = \log{\left(x + 1 \right)}$ ; to find $\frac{d}{d x} g{\left(x \right)}$ :
  1. Let $u = x + 1$ .
  2. The derivative of $\log{\left(u \right)}$ is $\frac{1}{u}$ .
  3. Then, apply the chain rule. Multiply by $\frac{d}{d x} \left(x + 1\right)$ :
    1. Differentiate $x + 1$ term by term:
      1. Apply the power rule: $x$ goes to $1$
      2. The derivative of the constant $1$ is zero.
      The result is: $1$
    The result of the chain rule is:
    
    $\frac{1}{x + 1}$
  The result is: $\frac{x}{x + 1} + \log{\left(x + 1 \right)}$
2. The derivative of a constant times a function is the constant times the derivative of the function.
  1. Let $u = 4 x$ .
  2. The derivative of cosine is negative sine:
    
    $\frac{d}{d u} \cos{\left(u \right)} = - \sin{\left(u \right)}$
  3. Then, apply the chain rule. Multiply by $\frac{d}{d x} 4 x$ :
    1. The derivative of a constant times a function is the constant times the derivative of the function.
      1. Apply the power rule: $x$ goes to $1$
      So, the result is: $4$
    The result of the chain rule is:
    
    $- 4 \sin{\left(4 x \right)}$
  So, the result is: $4 \sin{\left(4 x \right)}$
The result is: $\frac{x}{x + 1} + \log{\left(x + 1 \right)} + 4 \sin{\left(4 x \right)}$
Now simplify:

$\frac{x + \left(x + 1\right) \left(\log{\left(x + 1 \right)} + 4 \sin{\left(4 x \right)}\right)}{x + 1}$

The answer is:

$\frac{x + \left(x + 1\right) \left(\log{\left(x + 1 \right)} + 4 \sin{\left(4 x \right)}\right)}{x + 1}$

The graph

Plot the graph f(x)

Plot the graph f'(x)

The first derivative [src]

               x               
4*sin(4*x) + ----- + log(x + 1)
             x + 1

$$\frac{x}{x + 1} + \log{\left(x + 1 \right)} + 4 \sin{\left(4 x \right)}$$

Simplify

The second derivative [src]

  2                      x    
----- + 16*cos(4*x) - --------
1 + x                        2
                      (1 + x)

$$- \frac{x}{\left(x + 1\right)^{2}} + 16 \cos{\left(4 x \right)} + \frac{2}{x + 1}$$

Simplify

The third derivative [src]

                  3         2*x   
-64*sin(4*x) - -------- + --------
                      2          3
               (1 + x)    (1 + x)

$$\frac{2 x}{\left(x + 1\right)^{3}} - 64 \sin{\left(4 x \right)} - \frac{3}{\left(x + 1\right)^{2}}$$

Simplify

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Derivative of x*ln(x+1)-cos4x

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The solution