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(3x-1)/(2x+1)^2

How to use it?
Derivative of:
Derivative of cot(x)*cos(x)
Derivative of (x-1)
Derivative of x*e^x^2
Derivative of x*e^(-x^2)
Identical expressions
(3x- one)/(two x+ one)^2
(3x minus 1) divide by (2x plus 1) squared
(3x minus one) divide by (two x plus one) squared
(3x-1)/(2x+1)2
3x-1/2x+12
(3x-1)/(2x+1)²
(3x-1)/(2x+1) to the power of 2
3x-1/2x+1^2
(3x-1) divide by (2x+1)^2
Similar expressions
(3x+1)/(2x+1)^2
(3x-1)/(2x-1)^2

The derivative of the function/ (3x-1)/(2x+1)^2

Derivative of (3x-1)/(2x+1)^2

The solution

You have entered [src]

 3*x - 1  
----------
         2
(2*x + 1)

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

(3*x - 1)/(2*x + 1)^2

Detail solution

Apply the quotient rule, which is:

$\frac{d}{d x} \frac{f{\left(x \right)}}{g{\left(x \right)}} = \frac{- f{\left(x \right)} \frac{d}{d x} g{\left(x \right)} + g{\left(x \right)} \frac{d}{d x} f{\left(x \right)}}{g^{2}{\left(x \right)}}$

$f{\left(x \right)} = 3 x - 1$ and $g{\left(x \right)} = \left(2 x + 1\right)^{2}$ .

To find $\frac{d}{d x} f{\left(x \right)}$ :
1. Differentiate $3 x - 1$ term by term:
  1. The derivative of the constant $-1$ is zero.
  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: $3$
  The result is: $3$
To find $\frac{d}{d x} g{\left(x \right)}$ :
1. Let $u = 2 x + 1$ .
2. Apply the power rule: $u^{2}$ goes to $2 u$
3. Then, apply the chain rule. Multiply by $\frac{d}{d x} \left(2 x + 1\right)$ :
  1. Differentiate $2 x + 1$ term by term:
    1. The derivative of the constant $1$ is zero.
    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: $2$
    The result is: $2$
  The result of the chain rule is:
  
  $8 x + 4$
Now plug in to the quotient rule:

$\frac{3 \left(2 x + 1\right)^{2} - \left(3 x - 1\right) \left(8 x + 4\right)}{\left(2 x + 1\right)^{4}}$
Now simplify:

$\frac{7 - 6 x}{\left(2 x + 1\right)^{3}}$

The answer is:

$\frac{7 - 6 x}{\left(2 x + 1\right)^{3}}$

The graph

Plot the graph f(x)

Plot the graph f'(x)

The first derivative [src]

    3        (-4 - 8*x)*(3*x - 1)
---------- + --------------------
         2                 4     
(2*x + 1)         (2*x + 1)

$$\frac{\left(- 8 x - 4\right) \left(3 x - 1\right)}{\left(2 x + 1\right)^{4}} + \frac{3}{\left(2 x + 1\right)^{2}}$$

Simplify

The second derivative [src]

   /     -1 + 3*x\
24*|-1 + --------|
   \     1 + 2*x /
------------------
             3    
    (1 + 2*x)

$$\frac{24 \left(-1 + \frac{3 x - 1}{2 x + 1}\right)}{\left(2 x + 1\right)^{3}}$$

Simplify

The third derivative [src]

   /    8*(-1 + 3*x)\
24*|9 - ------------|
   \      1 + 2*x   /
---------------------
               4     
      (1 + 2*x)

$$\frac{24 \left(9 - \frac{8 \left(3 x - 1\right)}{2 x + 1}\right)}{\left(2 x + 1\right)^{4}}$$

Simplify

The graph

Derivative of (3x-1)/(2x+1)^2