[rant]
Has anyone at all ever had to find the second derivative of a function after that one college class you took but barely remember? I cannot possibly see how this:
y=(x-1)/(x-2) tan
3,2)
f'(3)=lim h->0 [((3+h-1)/(3+h-2))-(2/1)]/h
f'(3)=lim h->0 [(2+h-2(1+h))/(1+h)]/h
f'(3)= lim h->0 -h/(h(1+h))
f'(3)= -1
given that the tangent line to the curve y=f(x)at the point P(a, f(a)) is the line through P with slope M=lim a->0 (f(a+h)-f(a))/h
could possibly apply in any sort of real world application unless you want to be a math teacher.
[/rant]
Has anyone at all ever had to find the second derivative of a function after that one college class you took but barely remember? I cannot possibly see how this:
y=(x-1)/(x-2) tan
3,2) f'(3)=lim h->0 [((3+h-1)/(3+h-2))-(2/1)]/h
f'(3)=lim h->0 [(2+h-2(1+h))/(1+h)]/h
f'(3)= lim h->0 -h/(h(1+h))
f'(3)= -1
given that the tangent line to the curve y=f(x)at the point P(a, f(a)) is the line through P with slope M=lim a->0 (f(a+h)-f(a))/h
could possibly apply in any sort of real world application unless you want to be a math teacher.
[/rant]
