1/X by first principles
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
Where did -h come from
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
Oh, if you mean the algebra look at the numerator
[ x -(x+h)] = x - x - h
= -h
To find the derivative of 1/x using first principles (also known as the definition of the derivative), we need to follow these steps:
Step 1: Begin with the definition of the derivative:
The derivative of a function f(x) is given by the limit as h approaches 0 of [f(x + h) - f(x)] / h.
Step 2: Substitute the function f(x) with 1/x:
In this case, we want to find the derivative of f(x) = 1/x.
Step 3: Apply the definition of the derivative:
Now we substitute f(x) into the definition of the derivative:
f'(x) = lim(h -> 0) [(1/(x + h)) - (1/x)] / h.
Step 4: Simplify the expression:
To simplify the expression, we need to find a common denominator for the two terms in the numerator. The common denominator will be x * (x + h).
[1/(x + h) - 1/x] / h = [(x - (x + h)) / (x * (x + h))] / h
= [(x - x - h) / (x * (x + h))] / h
= [(-h) / (x * (x + h))] / h
= -1 / (x * (x + h))
Step 5: Take the limit as h approaches 0:
Now, we substitute h = 0 into the expression:
f'(x) = lim(h -> 0) [-1 / (x * (x + h))]
= -1 / (x * x)
= -1 / x^2
So, using first principles, we have found that the derivative of 1/x is -1 / x^2.