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

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Answer 1 · 2017-10-01T10:42:51

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

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.!