How do you find the derivative using limits of f(x)=4/sqrtx?

1 Answer
Oct 1, 2017

f'(x)=-2*x^(-3/2); x >0.

Explanation:

The Derivative of a Function f : RR to RR at a point x in RR, is

denoted by f'(x), and, is defined by,

lim_(t to x) (f(t)-f(x))/(t-x), if, the Limit exists.

We have, f(x)=4/sqrtx rArr f(t)=4/sqrtt.

Hence, (f(t)-f(x))/(t-x)=(4/sqrtt-4/sqrtx)/(t-x)={4(sqrtx-sqrtt)}/{sqrtt*sqrtx*(t-x)}.

:. lim_(t to x)(f(t)-f(x))/(t-x)=lim_(t to x){4(sqrtx-sqrtt)}/{sqrtt*sqrtx*(t-x)},

=lim{-4(sqrtt-sqrtx)}/{sqrtt*sqrtx*(t-x)}xx(sqrtt+sqrtx)/(sqrtt+sqrtx),

=lim(-4cancel((t-x)))/{sqrtt*sqrtx*cancel((t-x))*(sqrtt+sqrtx)},

=lim_(t to x)-4/{sqrtt*sqrtx*(sqrtt+sqrtx)},

=-4/{sqrtx*sqrtx*(sqrtx+sqrtx)},

=-4/(x*2sqrtx).

Thus, the Limit exists, and, therefore, for f(x)=4/sqrtx, we have,

f'(x)=-2*x^(-3/2); x >0..

Enjoy Maths.!