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Derivative of (2*x^3+1)/(x-4)

Function f() - derivative -N order at the point
v

The graph:

from to

Piecewise:

The solution

You have entered [src]
   3    
2*x  + 1
--------
 x - 4  
$$\frac{2 x^{3} + 1}{x - 4}$$
(2*x^3 + 1)/(x - 4)
Detail solution
  1. Apply the quotient rule, which is:

    and .

    To find :

    1. Differentiate term by term:

      1. The derivative of the constant 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: goes to

        So, the result is:

      The result is:

    To find :

    1. Differentiate term by term:

      1. The derivative of the constant is zero.

      2. Apply the power rule: goes to

      The result is:

    Now plug in to the quotient rule:


The answer is:

The graph
The first derivative [src]
     3           2
  2*x  + 1    6*x 
- -------- + -----
         2   x - 4
  (x - 4)         
$$\frac{6 x^{2}}{x - 4} - \frac{2 x^{3} + 1}{\left(x - 4\right)^{2}}$$
The second derivative [src]
  /              3       2 \
  |       1 + 2*x     6*x  |
2*|6*x + --------- - ------|
  |              2   -4 + x|
  \      (-4 + x)          /
----------------------------
           -4 + x           
$$\frac{2 \left(- \frac{6 x^{2}}{x - 4} + 6 x + \frac{2 x^{3} + 1}{\left(x - 4\right)^{2}}\right)}{x - 4}$$
The third derivative [src]
  /            3                  2  \
  |     1 + 2*x     6*x        6*x   |
6*|2 - --------- - ------ + ---------|
  |            3   -4 + x           2|
  \    (-4 + x)             (-4 + x) /
--------------------------------------
                -4 + x                
$$\frac{6 \left(\frac{6 x^{2}}{\left(x - 4\right)^{2}} - \frac{6 x}{x - 4} + 2 - \frac{2 x^{3} + 1}{\left(x - 4\right)^{3}}\right)}{x - 4}$$