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

2 Answers
maganbhai P.
Aug 14, 2018

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

Explanation:

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

MattyMatty
Aug 14, 2018

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

Explanation:

"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

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