Question

find the derivate of f(x)= square root(x-5)
using the tangent definition.

Answer 1

what is a tangent definition?

Answer 2

if you mean limits ... use a conjugate

Answer 3

yep
limits

Answer 4

with \[\lim_{x \rightarrow 0}\]

Answer 5

\[\lim\frac{f(x+h)-f(x)}{h}\]
\[\lim\frac{f(x+h)-f(x)}{h}\frac{f(x+h)+f(x)}{f(x+h)+f(x)}\]
\[\lim\frac{1}{h(f(x+h)+f(x))}\]
\[\lim\frac{1}{f(x+h)+f(x)}\]
\[\frac{1}{f(x+0)+f(x)}\]
\[\frac{1}{2f(x)}\]

Answer 6

the nature of a sqrt allows that to happen

Answer 7

got a typo after the conjugate :) should be an h to cancel, not a 1

Answer 8

i see

Answer 9

thank u very much
i got it now

Answer 10

....just for clean up
\[\lim\frac{f(x+h)-f(x)}{h}\frac{f(x+h)+f(x)}{f(x+h)+f(x)}\]
\[\lim\frac{f^2(x+h)-f^2(x)}{h[f(x+h)+f(x)]}\]
\[since~f(x) = \sqrt{(x-5)}\]
\[\lim\frac{h}{h[f(x+h)+f(x)]}\]
\[\lim\frac{1}{f(x+h)+f(x)}\]
etc ...

Answer 11

n the answer is wat i told u first 1/(power of function) *function with (power -1)*its deriviate
pwer is 2 function is x-5 ndits deriviate is 1. so end result is 1/(2(x-5)^(1/2))