How do you use the definition of a derivative to find the derivative of (x^2+1) / (x-2)?

1 Answer
Steve M
Mar 7, 2017

d/dx (x^2+1)/(x-2) = (x^2-4x-1)/(x-2)^2

Explanation:

The definition of the derivative of y=f(x) is:

f'(x)=lim_(h rarr 0) ( f(x+h)-f(x) ) / h

Let f(x) = (x^2+1)/(x-2) then, we will focus on the numerator f(x+h)-f(x) in the above limit definition;

f(x+h)-f(x)
\ \ = ((x+h)^2+1)/((x+h)-2) - (x^2+1)/(x-2)

\ \ = { ((x+h)^2+1)(x-2) - (x^2+1)(x+h-2) } / { (x+h-2)(x-2) }

\ \ = { (x^2+2hx+h^2+1)(x-2) - (x^2+1)(x+h-2) } / { (x+h-2)(x-2) }

\ \ = {{ x^3+2hx^2+h^2x+x -2x^2-4hx-2h^2-2 -x-h+2-x^3-hx^2+2x^2 }} / { (x+h-2)(x-2) }

\ \ = { hx^2+h^2x -4hx-2h^2 -h } / { (x+h-2)(x-2) }

And so the limit becomes:

f'(x)=lim_(h rarr 0) {{ hx^2+h^2x -4hx-2h^2 -h } / { (x+h-2)(x-2) }}/h

" "= lim_(h rarr 0) { x^2+hx -4x-2h -1 } / { (x+h-2)(x-2) }

" "= { x^2+0 -4x-0 -1 } / { (x+0-2)(x-2) }

" "= { x^2-4x-1 } / { (x-2)(x-2) }

" "= { x^2-4x-1 } / { (x-2)^2 }

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