Answer to Question #229660 in Calculus for Guddy

Question

By the first principle of differentiation,find the derivative of f(X)=2/2x-1 at X=1

Answer 1

Let us find the derivative of $f(x)=\frac{2}{2x-1}$ at $x=1$ using the first principle of differentiation:

$f'(1)=\lim\limits_{x\to 1}\frac{f(x)-f(1)}{x-1} =\lim\limits_{x\to 1}\frac{\frac{2}{2x-1}-\frac{2}{2\cdot 1-1}}{x-1} =\lim\limits_{x\to 1}\frac{\frac{2}{2x-1}-2}{x-1} =\lim\limits_{x\to 1}\frac{\frac{2-4x+2}{2x-1}}{x-1}\\ =\lim\limits_{x\to 1}\frac{4-4x}{(2x-1)(x-1)} =\lim\limits_{x\to 1}\frac{-4(x-1)}{(2x-1)(x-1)} =\lim\limits_{x\to 1}\frac{-4}{2x-1} =\frac{-4}{2\cdot 1-1}=-4.$