Find the derivative of the function f(x) = (x | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $f(x) = (x-1)(x-2) = x^2 - 3x + 2$. By the first principle of derivatives,$f'(x) = \lim_{h 	o 0} \frac{f(x+h) - f(x)}{h}$. Substituting $f(x+

Vedclass · Accepted Answer

$2x - 3$

Vedclass · Answer

(A) Let $f(x) = (x-1)(x-2) = x^2 - 3x + 2$. By the first principle of derivatives,$f'(x) = \lim_{h 	o 0} \frac{f(x+h) - f(x)}{h}$. Substituting $f(x+h) = (x+h)^2 - 3(x+h) + 2$: $f'(x) = \lim_{h 	o 0} \frac{[(x+h)^2 - 3(x+h) + 2] - [x^2 - 3x + 2]}{h}$ $f'(x) = \lim_{h 	o 0} \frac{x^2 + 2xh + h^2 - 3x - 3h + 2 - x^2 + 3x - 2}{h}$ $f'(x) = \lim_{h 	o 0} \frac{2xh + h^2 - 3h}{h}$ $f'(x) = \lim_{h 	o 0} (2x + h - 3)$ As $h 	o 0$,$f'(x) = 2x - 3$.

Find the derivative of the function $f(x) = (x-1)(x-2)$ using the first principle.

Similar Questions

Find the derivative of $f$ from the first principle,where $f$ is given by $f(x) = \frac{2x+3}{x-2}$.

Find the derivative of the function $f(x) = \frac{x+1}{x-1}$ using the first principle.

Find the derivative of $\sin x$ at $x=0$.

Compute the derivative of $\tan x$.

If $f(r) = \pi r^2$,then $\lim_{h \to 0} \frac{f(r + h) - f(r)}{h} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module