Using a limit of slopes, find the slope of the tangent line to y=(1/x) at x=c.?

Question

a nice detailed procedure to answer this starting with y2-y1/x2-x1 would be awesome. at least i think thats the first step...right?

Answer 1

Judging from the wording of your question, my guess is that you haven't proven the power rule for derivatives in class yet, so the above poster's answer is useless to you. Fortunately, the limit is not hard. Consider the following:

[x→c]lim (1/x - 1/c)/(x-c)
[x→c]lim (c/(cx) - x/(cx))/(x-c)
[x→c]lim (c-x)/(cx(x-c))
[x→c]lim -1/(cx)
-1/c²

And we are done.

Answer 2

? if P=(x1, y1) and Q=(x2,y2) are distinctive factors then you are able to draw a directly line PQ via them: y-y1 =ok*(x-x1), the place ok=(y2-y1)/(x2-x1) is suggested because of the fact the slope of the line PQ; ? if component P on our line has x=c, then y=2c^2 -c, it somewhat is P=(c, 2c^2 –c); ? if component Q on our line has x=b, then y=2b^2 -b, it somewhat is Q=(b, 2b^2 –b); ? consequently the slope of the line PQ in our case is ok=((2b^2 –b) –(2c^2 –c)) /(b-c) = = (2(b^2 –c^2) –(b-c)) /(b-c) = 2b+2c -a million; and PQ isn't yet tangent! ? if distance between P and Q is sufficiently small, you are able to say that PQ is tangent to our line y=2x^2 –x at component P; and sufficiently small ability b ? c; then ok=2c +2c –a million =4c –a million is the slope in question;

Answer 3

to get the slope of the line tangent to any point in a curve, get the curve's derivative

y = (1/x) = x^-1

dy/dx = -1 x^-2
slope = dy/dx = -1/x^2

at x=c,
slope = -1 / c^2