find from first principles the derivative of y=1/x?

Answer 1

First principles, I assume, mean from the definition of derivative as the instantaneous rate of change. That is, the limit as n ---> 0 of ((y(x)-y(x+n))/(x-(x+n)).

So, plugging in that y = 1/x, we get the limit as n ---> 0 of (1/x - 1/(x+n))/-n. We want to get this into a form where we can substitute 0 for n throughout the equation and get a result. We can't do that now because n is the denominator of a fraction and making the substitution would mean dividing by 0.

My first instinct here would be to find the LCD for 1/x and 1/(x+n). We can rewrite these as (x+n)/(x^2 + nx) and x/(x^2 + nx). Then our limit becomes the limit as n ---> 0 of (n/(x^2 + xn))/-n.

The next thing I see is that we have a fraction as the numerator in another fraction. We can simplify that. Instead of dividing by -n, we can multiply by 1/-n and get the limit as n ---> 0 of n/-(nx^2 + xn^2).

Now, each term has at least one n in it, so divide each term by n and you get the limit as n ---> 0 of -1/(x^2 + xn).

Now we can substitute in 0 for n and find that the limit is -1/(x^2 + 0x), or -1/x^2. This is the derivative of 1/x.

Checking the answer with calculus: y = 1/x = x^-1. Power Rule: y' = -1x^-2 or -1/x^2.

Answer 2

I don't know what you mean by first principle, I guess it is the definition of derivative. If I'm right, we have to compute

y' = lim h --> 0 (1/(x + h) - 1/x)/h,

For h <>0 such that x + h <>0, we have

(1/(x + h) - 1/x)/h = (-h/(x(x +h))/ h = -1/(x(x+h)). So, when h --> 0, this expression tends to -1h/(x*x) = -1/x^2

It follows that y' = -1/x^2, for x <>0.

Of course thia cab be done by the general rule for differentiating a power function y = x^a, that is, y' = a x^(a -1).
Putting a = -1, we get y' = - x^(-2) = -1/x^2
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Answer 3

not easy without a diagram - but here goes:

the derivative is basically the gradient of the curve at point (x,y) or (x,1/x).

take another point near (x,y) and call it (x+h, 1/(x+h))

the slope between these two points is:

{1/(x+h)-1/x}/h difference in y values/difference in x values

simplify to

-1/(x^2+xh)

the derivative or the gradient of the curve at the point (x,y) is when h->0. In other words when the distance between the two points is negligible.

hence the gradient or derivative is -1/x^2

Answer 4

y=x^-1 power rule y'= -1/y^2

Answer 5

f `(x) = lim h->0 [ f(x + h) - f(x) ] / h ----h ≠ 0
f `(x) = lim h->0 1 / [ 1/(x + h) - 1/x) ] / h
f `(x) = lim h->0 [x - (x + h) ] / hx(x + h)
f `(x) = lim h->0 [- h] / (h)(x)(x + h)
f `(x) = - 1 / x²

dy/dx = f `(x) = - 1 / x²