Asked by Lilz
                find from first principle,the derivative of 2x^2-x with respect to x
            
            
        Answers
                    Answered by
            Anonymous
            
    y(x) = 2 x^2 - x
y (x+h) = 2 (x+h)^2 - x - h
so
y(x+h) = 2 x^2 + 4 x h + 2 h^2 - x - h
and
y(x+h)-y(x) = 4 x h + 2 h^2
divide by h
[y(x+h)-y(x)] / h = 4 x + 2 h
let h ---> 0
4 x
    
y (x+h) = 2 (x+h)^2 - x - h
so
y(x+h) = 2 x^2 + 4 x h + 2 h^2 - x - h
and
y(x+h)-y(x) = 4 x h + 2 h^2
divide by h
[y(x+h)-y(x)] / h = 4 x + 2 h
let h ---> 0
4 x
                    Answered by
            Anonymous
            
    TYPO !
y(x) = 2 x^2 - x
y (x+h) = 2 (x+h)^2 - x - h
so
y(x+h) = 2 x^2 + 4 x h + 2 h^2 - x - h
and
y(x+h)-y(x) = 4 x h + 2 h^2 - h
divide by h
[y(x+h)-y(x)] / h = 4 x + 2 h -1
let h ---> 0
4 x - 1
    
y(x) = 2 x^2 - x
y (x+h) = 2 (x+h)^2 - x - h
so
y(x+h) = 2 x^2 + 4 x h + 2 h^2 - x - h
and
y(x+h)-y(x) = 4 x h + 2 h^2 - h
divide by h
[y(x+h)-y(x)] / h = 4 x + 2 h -1
let h ---> 0
4 x - 1
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