Asked by Kristin
1/X by first principles
Answers
Answered by
Damon
What do you want, the derivative?
if so
f(x) = 1/x
f(x+h) = 1/(x+h)
f(x+h)-f(x) = 1/(x+h) - 1/x
= [ x -(x+h)] /[x(x+h)]
= -h/ (x^2 + x h)
[f(x+h)-f(x)]/h = -1/(x^2+xh)
limit as h-->0
= -1/x^2
which is of course the derivative
if so
f(x) = 1/x
f(x+h) = 1/(x+h)
f(x+h)-f(x) = 1/(x+h) - 1/x
= [ x -(x+h)] /[x(x+h)]
= -h/ (x^2 + x h)
[f(x+h)-f(x)]/h = -1/(x^2+xh)
limit as h-->0
= -1/x^2
which is of course the derivative
Answered by
Kristin
Where did -h come from
Answered by
Damon
h is a small increase in x which is allowed to approach zero to get the slope of f(x) at x, the definition of the derivative
Definition of deriviative of f(x) is:
df/dx = {f(x+h) - f(x) ]/h as h-->0
Definition of deriviative of f(x) is:
df/dx = {f(x+h) - f(x) ]/h as h-->0
Answered by
Damon
Oh, if you mean the algebra look at the numerator
[ x -(x+h)] = x - x - h
= -h
[ x -(x+h)] = x - x - h
= -h
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.