# Find the derivative of the function f(x)=2x^2-x+1 from first principles.

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### 4 Answers

To find derivative from first principle the steps followed are given below:

- Find value of (y + dy) as f(x+dx)
- Find value of dy = f(x+dx) - f(x)
- Find value of dy/dx

Given:

y = f(x) = 2*x^2 - x +1

Therefore: y + dy = f(x+dx) = 2*(x+dx)^2 - (x+dx) +1

= 2*(x^2 + 2x*dx + dx^2) - x - dx + 1

= 2*x^2 + 4x*dx + 2*dx^2) - x - dx + 1

And dy = f(x+dx) - f(x)

= 2*x^2 + 4x*dx + 2*dx^2) - x - dx + 1 - [2*x^2 - x +1]

= 4x*dx + 2*dx^2 - dx

Then: dy/dx = [- 4x*dx + 2*dx^2 - dx]/dx

= 4x + 2dx -1

When dx approaches zero, 2dx approaches zero

There Limit dy/dx when sx approaches zero is: 4x - 1

Thus derivative of f(x) = 2*x^2 - x +1 is

= 4x - 1

f(x) = 2x^2-x+ 1

The derivative of f(x) is defined as :

d/dx (f(x) )= f'(x) = Limit f(x+deltax))-f(x).)/deltax as deltax tends to zero. Call deltax = h for convenience of writing.

Then d/dx(f(x)) = d/dx ({2x^2-x+1}) = f'(x) = limt[f(x+h-f(x)]/h as h -->0

=Limit{[2(x+h)^2-(x+h)+1]-[2x^2-x+1]}/h as h-->0

=Limit{2(x^2+2hx+h^2 -(x+h)+1 )-2x^2 +x -1}h as h-->

=Limit{2x^2+4hx+h^2-x-h-2x^2+x-1}/h as h-->0

=Limit{0*x^2+4hx-h+h^2}/h as h-->0

=Limit {4hx/h -h/h+h} as h-->0

=Limit 4x-1 +h as h-->0

=4x-1+0

=4x -1.

Therefore, f'(x) = d/dx(2x^2-x+1) = 4x-1 derived from first principles.

We'll consider the function above as a sum of the following elementary functions:

- (2x^2) - is a power function and it's derivtaive will be calculated with the rule of power functions. We know that (x^n) is a power function type, and it's derivative is (x^n)'=n*[x^(n-1)].We'll do the same way with (2x^2), where (2x^2)'=2*2*x^(2-1)=4x;
- (-x) - is a linear function, where it's first derivative is (-x)'=-1. (-x) can be as well considered as a power function where the power of the unknown x, is equal to 1. In each power function , the rule of finding out the derivative is: (x^n)=x*x*x*...*x. As we can se. it's all about a product of n factors. The rule in this case is: (x^n)'=x'*x^(n-1)+x*x'*x*...+....x*x*...*x'=n*x^(n-1) So,(-x)'=(-1)'*x+(-1)*(x)'=0*x+(-1)*1=-1
- (+1)- is a constant function, where it's derivative is 0.

**f'(x)= (2x^2-x+1)'=(2x^2)'+(-x)'+(1)'=4x-1**

f(x)=2x^2-x+1

f'(x) = 4x - 1

You have to apply the chain rule for derivatives.