differentiate this using first principles.?

Question

y = 1 / (x^2) using first principles.

Answer 1

Let:

y = 1 / (x^2) = f(x).

Therefore, f(x + h) - f(x) = 1/(x + h)^2 -1/x^2

= [(x^2) - (x + h)^2] / [x^2*(x + h)^2]. Expanding & simplifying

f(x + h) - f(x) = [-2x*h - h^2]/[x^2*(x + h)^2]. Dividing both sides by non-zero 'h'

(1/h)*[f(x+h) - f(x)] = [- 2x - h]/[x^2*(x + h)^2].

Now taking limits on both sides as h approaches zero:

f'(x) or dy/dx = -2x/(x^4) = -2/(x^3).

Answer 2

Differentiation from first principles means using the basic definition of dy/dx which is
dy/dx = limit as h approaches 0 of (f(x + h) - f(x))/h.

Therefore you need to work on 1/((x + h)^2) - 1/(x^2). You will need to carefully combine these into one fraction first. Of course the power rule will tell you what the last line must be.

Take it from there and try to do it yourself before looking at any detailed answer left here.

Answer 3

i would be abbreviating notation, only assume I mean lim as h is going to 0 as quickly as I write lim and etc. f'(x) = lim (tan(2(x+h)) - tan 2x) / h f'(x) = lim (sin(2(x+h))/cos(2(x+h)) - sin2x/cos2x) / h f'(x) = lim ((sin 2x cos 2h + sin 2h cos 2x) / cos(2(x+h))) - sin 2x / cos 2x) / h Now i'm breaking it up into 2 limits. f'(x) = lim sin 2h cos 2x / (h cos (2(x+h))) + lim ((sin 2x cos 2h) / cos (2(x+h)) - sin 2x / cos 2x) / h ok. The left shrink is going to 2 cos 2x / cos 2x = 2, definitely. The sin 2h/h is going to 2, and the cos (2(x+h)) is going to cos 2x, so it cancels with the cos 2x interior the numerator. So all of it is going to easily 2. As for the excellent suited shrink, i visit multiply the left facet with the aid of cos 2x / cos 2x and the excellent suited facet with the aid of cos (2(x+h)) / cos (2(x+h)). this variety i can deliver each little thing into one denominator and then magic happens. f'(x) = 2 + lim (sin 2x cos 2h cos 2x - sin 2x cos (2(x+h))) / (h cos 2x cos (2x + 2h)) with a bit of luck that final step isn't perplexing to you. I only positioned each little thing on a similar denominator with the aid of multiplying with the aid of cos 2x/ cos 2x or cos (2(x+h)) / cos (2(x+h)) the place ideal. Now we element our sin 2x from the numerator. f'(x) = 2 + lim (sin 2x (cos 2h cos 2x - cos (2(x+h))) / (h cos 2x cos (2x + 2h))) next I chop up up the cos (2x + 2h) using the identity cos(x + y) = cos x cos y - sin x sin y f'(x) = 2 + lim (sin 2x (cos 2h cos 2x - (cos 2h cos 2x - sin 2x sin 2h)) / (h cos 2x cos (2x + 2h))) f'(x) = 2 + lim (sin 2x sin 2x sin 2h / (h cos 2x cos (2x + 2h))) f'(x) = 2 + lim (sin 2x)^2 * sin 2h / (h cos 2x cos (2x + 2h)) The sin 2h/h is going to 2, and something is going to (sin 2x)^2 / (cos 2x)^2. The cos (2x + 2h) only is going to cos 2x. So the full shrink is going to 2 * (sin 2x)^2 / (cos 2x)^2 = 2 (tan 2x)^2 f'(x) = 2 + 2tan(2x)^2 f'(x) = 2(a million + (tan 2x)^2) = 2 (sec 2x)^2 it rather is the envisioned effect.