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The derivative of the function/ y=x\x-2x^2+1

You entered:

y=x\x-2x^2+1

What you mean?

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

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

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

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

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

x/x - 2*x^2 + 1

x/x - 2*x^2 + 1

Derivative of y=x\x-2x^2+1

The solution

You have entered [src]

x      2    
- - 2*x  + 1
x

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

d /x      2    \
--|- - 2*x  + 1|
dx\x           /

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

Transform

Detail solution

Differentiate $- 2 x^{2} + 1 + \frac{x}{x}$ term by term:
1. Apply the quotient rule, which is:
  
  $\frac{d}{d x} \frac{f{\left(x \right)}}{g{\left(x \right)}} = \frac{- f{\left(x \right)} \frac{d}{d x} g{\left(x \right)} + g{\left(x \right)} \frac{d}{d x} f{\left(x \right)}}{g^{2}{\left(x \right)}}$
  
  $f{\left(x \right)} = x$ and $g{\left(x \right)} = x$ .
  
  To find $\frac{d}{d x} f{\left(x \right)}$ :
  1. Apply the power rule: $x$ goes to $1$
  To find $\frac{d}{d x} g{\left(x \right)}$ :
  1. Apply the power rule: $x$ goes to $1$
  Now plug in to the quotient rule:
  
  $0$
2. The derivative of a constant times a function is the constant times the derivative of the function.
  1. The derivative of a constant times a function is the constant times the derivative of the function.
    1. Apply the power rule: $x^{2}$ goes to $2 x$
    So, the result is: $4 x$
  So, the result is: $- 4 x$
3. The derivative of the constant $1$ is zero.
The result is: $- 4 x$

The answer is:

$- 4 x$

The graph

Plot the graph f(x)

Plot the graph f'(x)

The first derivative [src]

-4*x

$$- 4 x$$

Simplify

The second derivative [src]

-4

$$-4$$

Simplify

The third derivative [src]

$$0$$

Simplify

The graph

$Derivative of y=x\x-2x^2+1$