Derivative of ln(x²+x-1)

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   / 2        \
log\x  + x - 1/

\log{\left(\left(x^{2} + x\right) - 1 \right)}

log(x^2 + x - 1)

Detail solution

Let $u = \left(x^{2} + x\right) - 1$ .
The derivative of $\log{\left(u \right)}$ is $\frac{1}{u}$ .
Then, apply the chain rule. Multiply by $\frac{d}{d x} \left(\left(x^{2} + x\right) - 1\right)$ :
1. Differentiate $\left(x^{2} + x\right) - 1$ term by term:
  1. Differentiate $x^{2} + x$ term by term:
    1. Apply the power rule: $x^{2}$ goes to $2 x$
    2. Apply the power rule: $x$ goes to $1$
    The result is: $2 x + 1$
  2. The derivative of the constant $-1$ is zero.
  The result is: $2 x + 1$
The result of the chain rule is:

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

$\frac{2 x + 1}{x^{2} + x - 1}$

The answer is:

\frac{2 x + 1}{x^{2} + x - 1}

The graph

Plot the graph f(x)

Plot the graph f'(x)

The first derivative [src]

 1 + 2*x  
----------
 2        
x  + x - 1

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

Simplify

The second derivative [src]

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

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

Simplify

The third derivative [src]

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

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

Simplify

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Derivative of ln(x²+x-1)

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Piecewise:

The solution