How do you use the definition of a derivative to find the derivative of #f(x)=x^3-2x^2+5x-6#?

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Answer 1 · 2018-08-14T18:37:43

#f'(x)=3x^2-4x+5#

Here ,

#f(x)=x^3-2x^2+5x-6=>f(t)=t^3-2t^2+5t-6#

We know that ,

#f'(x)=lim_(t tox)(f(t)-f(x))/(t-x) to"definition"#

Substitute values of #f(t) and f(x)#

#f'(x)=lim_(t tox)((t^3-2t^2+5t-6)-(x^3-2x^2+5x-6))/(t-x)#

#f'(x)=lim_(t tox)((t^3-x^3)-2(t^2-x^2)+5(t-x))/(t-x)#

#f'(x)=lim_(t tox)(color(red)(cancel((t-x))){(t^2+tx+x^2)-2(t+x)+5(1)})/color(red)cancel(t-x)#

#f'(x)=lim_(t tox){(t^2+tx+x^2)-2(t+x)+5(1)}#

#f'(x)=(x^2+x*x+x^2)-2(x+x)+5(1)#

#f'(x)=(3x^2)-2(2x)+5(1)#

#:.f'(x)=3x^2-4x+5#

Answer 2 · 2018-08-14T19:26:24

#f'(x) = 3 x^2 - 4 x + 5#

#"The definition of the derivative of "f(x)" is : "#

#f'(x) = lim_{h->0} ( f(x+h) - f(x) ) / h#

#"So here we have "#

#lim_{h->0} ( (x+h)^3 - 2 (x+h)^2 + 5 (x+h) - 6 - x^3 + 2 x^2 - 5 x + 6 ) / h#

#= ((x^3 + 3 h x^2 + 3 h^2 x + h^3) - 2 (x^2 + 2 h x + h^2) + 5 (x + h) - x^3 + 2 x^2 - 5 x)/h#

#= lim_{h->0} (3 h x^2 + 3 h^2 x + h^3 - 4 h x - 2 h^2 + 5 h)/h#

#= lim_{h->0} 3 x^2 + 3 h x + h^2 - 4 x - 2 h + 5#

#= lim_{h->0} 3 x^2 - 4 x + 5 + h^2 + 3 h x - 2 h#

#= 3 x^2 - 4 x + 5 + lim_{h->0} h^2 + 3 h x - 2 h#

#= 3 x^2 - 4 x + 5 + 0#

#= 3 x^2 - 4 x + 5#