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

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

Question

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

Vedclass · Accepted Answer

$\frac{-2}{(x-1)^{2}}$

Vedclass · Answer

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

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

Similar Questions

Compute the derivative of $\tan x$.

Find the derivative of the function $f(x) = \frac{1}{x^{2}}$ using the first principle.

$\mathop {\lim }\limits_{h \to 0} \frac{{\sqrt {x + h} - \sqrt x }}{h} = $

Find the derivative of $f(x) = x^2$.

$\mathop {\lim }\limits_{y \to 0} \frac{(x + y)\sec (x + y) - x\sec x}{y} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module