if f(x)=1/x then f'(x)= -1/x^2 need help on derivative?

Question

i need to know how to get -1/x^2 using the lim h ->0
(f(x+h)-f(x))/ h (derivative function)
i tried couple of times but it's frustrating with the triple fractions i can't seem to get the answer.
someone help~

i asked by using the derivative function.

all the answers i got.. all so confusing. =[

THANKYOU PASCAL!!!

Answer 1

First, write the definition:

[h→0]lim (1/(x+h) - 1/x)/h

Find a common denominator:

[h→0]lim (x/(x(x+h)) - (x+h)/(x(x+h)))/h

Add:

[h→0]lim (-h/(x(x+h)))/h

Cancel the h's in the numerator and denominator:

[h→0]lim -1/(x(x+h))

Take the limit:

-1/(x(x+0))
-1/x²

And we are done.

Answer 2

Well, to start off with, ln(x^2+2) is not ln(x^2) * ln(2). The basic logarithmic equation is ln(xy) = ln(x) + ln(y). There's no simplification you can make for ln(x+y). With that out of the way, using the chain rule and the fact that the derivative of ln(x) is 1/x you get f'(x) = 1/(x^2+2) * d/dx(x^2+2) = 2x/(x^2+2) and f''(x) will be the derivative of that function, which you can work out by the quotient rule and gives you f"(x) = -2(x^2 - 2) / (x^2 + 2)^2

Answer 3

This is rather cumbersome to write out in this format but I'll try my best. Hopefully this will be understandable:

We are taking the limit as h --> 0 of
[(1/x+h) - (1/x)]/h

Rewrite as
1/h(x+h) - 1/hx

Now create a common denominator, which gives us
hx/[hx(h(x+h))] - h(x+h)/[hx(h(x+h))]

Combine
(hx - h(x+h))/[hx(h(x+h))]

Simplify
-h^2/[hx(hx+h^2)]

Cancel the h's
-h/x(hx+h^2)

-h/(hx^2 + h^2(x))

Cancel another set of h's (You could have done this all in one step, sorry but this was how I originally did it)
-1/(x^2 + hx)

Now we are taking the limit of -1/(x^2 + hx) as h approaches 0.

Lim h--> 0 of -1/(x^2 + hx) = -1/x^2

which is indeed the derivative of 1/x.

Answer 4

you mean the three-step-rule of finding the derivative...

y = 1/x

step 1
delta y = 1/(x + delta x) - 1/x
combining the terms in the right side, first by getting the LCD,
= [ x - (x + delta x) ] / [ x(x + deltax) ]
= [ x - x - delta x) ] / [ x(x + deltax) ]
delta y = -delta x / [ x(x + deltax) ]

step 2 (divide delta y by delta x)
delta y / delta x = -1 / [ x(x + deltax) ]

step 3 (evaluate the f'(x) = lim of [delta y/delta x] as delta x approaches 0. we have..

f'(x) = -1 / [x(x+0)] = -1 / x^2

btw, i saw you wrote
lim h ->0 (f(x+h)-f(x))/ h

my solution might confuse you. i used delta x instead of h..it is just the same as you may see in some books.

Answer 5

to find the derivative to 1/x directly, rewrite this expession as

x^(-1); by exponent law:

(dy/dx) =
-1 * x^ (-2)
-1 / (x^2)

Answer 6

f `(x) = lim x->0 [ f(x + h) - f(x) ] / h = N / D
N = 1 / (x + h) - 1 / x
N = [ x - (x + h) ] / [ (x)(x + h) ]
N = - h / (x)(x + h)
N/D = - 1 / (x)(x + h)
N/D = - 1 / x ² as h ->0
f `(x) = - 1 / x ²

Answer 7

f(x) = 1/x
f(x) = x^-1
f'(x) = -1*( x^ (-1-1))
f'(x) = -1*(x^-2)
f'(x) = -1/x^2