ladyrae
Jun12-04, 03:01 PM
My algebra continues to let me down…
How about this one…I know there a simpler ways to do these problems but at this point I’m supposed to do it the hard way.
Using the definition of derivate find f ` (x)
f(x) = SQRT(1-3x)
f ` (x) = lim h->0 [(f(x+h)) – (f(x))]/h
= lim->0 [(SQRT(1-3(x+h))) – (SQRT(1-3x))]/h
= lim->0 [[(SQRT(1-3(x+h))) – (SQRT(1-3x))]/h] . [[((SQRT(1-3x-3h))) + (SQRT (1-3x))]/ [((SQRT(1-3x-3h))) + (SQRT (1-3x))]]
=lim->0 (1-3x-3h-1+3x)/(h[(SQRT(1-3x-3h)) + (SQRT (1-3x))]
= lim->0 -3/[SQRT(1-3x-3h) + SQRT (1-3x)]
= -3/(2(SQRT(1-3x))
How about this one…I know there a simpler ways to do these problems but at this point I’m supposed to do it the hard way.
Using the definition of derivate find f ` (x)
f(x) = SQRT(1-3x)
f ` (x) = lim h->0 [(f(x+h)) – (f(x))]/h
= lim->0 [(SQRT(1-3(x+h))) – (SQRT(1-3x))]/h
= lim->0 [[(SQRT(1-3(x+h))) – (SQRT(1-3x))]/h] . [[((SQRT(1-3x-3h))) + (SQRT (1-3x))]/ [((SQRT(1-3x-3h))) + (SQRT (1-3x))]]
=lim->0 (1-3x-3h-1+3x)/(h[(SQRT(1-3x-3h)) + (SQRT (1-3x))]
= lim->0 -3/[SQRT(1-3x-3h) + SQRT (1-3x)]
= -3/(2(SQRT(1-3x))