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y=e^[ln(1-2x+x²)]

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Identical expressions
y=e^[ln(one -2x+x²)]
y equally e to the power of [ln(1 minus 2x plus x²)]
y equally e to the power of [ln(one minus 2x plus x²)]
y=e[ln(1-2x+x²)]
y=e[ln1-2x+x²]
y=e^[ln1-2x+x²]
Similar expressions
y=e^[ln(1+2x+x²)]
y=e^[ln(1-2x-x²)]

The derivative of the function/ y=e^[ln(1-2x+x²)]

Derivative of y=e^[ln(1-2x+x²)]

The solution

You have entered [src]

    /           2\
 log\1 - 2*x + x /
E

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

E^log(1 - 2*x + x^2)

Detail solution

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

$2 x - 2$

The answer is:

$2 x - 2$

The graph

Plot the graph f(x)

Plot the graph f'(x)

The first derivative [src]

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

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

Simplify

The second derivative [src]

$$2$$

Simplify

The third derivative [src]

$$0$$

Simplify

The graph

Derivative of y=e^[ln(1-2x+x²)]