Compute derivative using the definition of the derivative.

Question

f(x)=1+1/x^2

1+ 1/(x+h)^2 - (1+ 1/(x^2))
------ divide by h and multiply by reciprocal

1/h(x+h)^2 - 1/hx^2

Next I expanded it and got this messy huge number and multiply the bottom of each other after cancelling some terms out

(2h^2x+h^3)/(h^2x^4+2h^3x^3+h^4x^2) That's as far as I got and I'm not quite sure what I did wrong or what to do next, however I checked online and the actual derivative is -2/x^3

Answer 1

f´(x)= lim [ f ( x + ∆h ) - f ( x ) ] / ∆h
∆h-> 0

In this case :

f ( x ) = 1 + 1 / x ^ 2

f ( x + ∆h ) = 1 + 1 / ( x + ∆h ) ^ 2

f ( x + ∆h ) - f ( x ) =

1 + 1 / ( x + ∆h ) ^ 2 - ( 1 + 1 / x ^ 2 ) =

1 + 1 / ( x + ∆h ) ^ 2 - 1 - 1 / x ^ 2 =

1 / ( x + ∆h ) ^ 2 - 1 / x ^ 2 =

x ^ 2 / [ x ^ 2 * ( x + ∆h ) ^ 2 ] - ( x + ∆h ) ^ 2 / [ x ^ 2 * ( x + ∆h ) ^ 2 ] =

[ x ^ 2 - ( x + ∆h ) ^ 2 ] / [ x ^ 2 * ( x + ∆h ) ^ 2 ] =

[ x ^ 2 - ( x ^ 2 + 2 x ∆h + ∆h ^ 2 ) ] / [ x ^ 2 * ( x + ∆h ) ^ 2 ] =

[ x ^ 2 - x ^ 2 - 2 x ∆h - ∆h ^ 2 ] / [ x ^ 2 * ( x + ∆h ) ^ 2 ] =

[ - 2 x ∆h - ∆h ^ 2 ] / [ x ^ 2 * ( x + ∆h ) ^ 2 ] =

∆h [ - 2 x - ∆h ] / [ x ^ 2 * ( x + ∆h ) ^ 2 ]

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

Now:

f´(x)= lim [ f ( x + ∆h ) - f ( x ) ] / ∆h
∆h-> 0

f´ (x)= lim { ∆h [ - 2 x - ∆h ] / [ x ^ 2 * ( x + ∆h ) ^ 2 ] } / ∆h
∆h-> 0

f ´ (x)= lim ∆h [ - 2 x - ∆h ] / { ∆h [ x ^ 2 * ( x + ∆h ) ^ 2 ] }
∆h-> 0

f ´ (x)= lim [ - 2 x - ∆h ] / [ x ^ 2 * ( x + ∆h ) ^ 2 ]
∆h-> 0

As ∆h-> 0 then:

- 2 x - ∆h = - 2 x - 0 = - 2 x

( x + ∆h ) ^ 2 = ( x + 0 ) ^ 2 = x ^ 2

so :

f´(x)= lim [ - 2 x - ∆h ] / [ x ^ 2 * ( x + ∆h ) ^ 2 ] =
∆h-> 0

- 2 x / ( x ^ 2 * x ^ 2 ) =

- 2 x / x ^ 4 =

- 2 x / ( x * x ^ 3 ) =

- 2 / x ^ 3

Answer 2

In your homework you can replace ∆h with h

Is all the same.