Solve derivative by using the definition of the derivative limit

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The discussion focuses on solving the derivative of the function h(x) = (x-1)-2 using the definition of the derivative limit. Participants detail the process of finding the derivative by manipulating the limit expression and simplifying the numerator. Key steps include expanding the numerator and canceling terms to resolve the limit as h approaches 0. The conversation highlights common pitfalls in algebraic manipulation and emphasizes the importance of correctly applying the limit definition.

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solve derivative by using the definition of the derivative limit ,,,

hi alll


1. Homework Statement [/b]

find the derivative by using the definition of the derivative limit

h(x)= (x-1)^-2




3. The Attempt at a Solution


well ,,, first

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

f(h+x) = 1/(x+h-1)^2 =====> 1/(x+h)^2-2hx+1 ===> 1/(x^2+2hx+h^2)-2hx+1


f(h+x)-f(x) = (1/x^2+h^2+1) - (1/x^2-2x+1) ===> x^2-2x+1-x^2+h^2+1/(x^2+h^2+1).(x^2-2x+1)


so it will be ,,,, -2x-h^2/x^4-2x^3+x^2 +(h^2.x^2)+2xh^2+h+x^2-2x+1


now adding the lim

lim(h=>0) (h^2+2x/x^4-2x^3+x^2 +(h^2.x^2)+2xh^2+h+x^2-2x+1)/h


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can someone tell me what did i do wrong so i can't divide the H down and the one up ,,,

i been trying to solve it from 2days ,,, =,=
 
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Uhh... let's ignore where the mistake was (the hell I'm going to try read through that mess) and will simply guide you through it instead.

So this is what we have,

\frac{\frac{1}{(x+h-1)^2}-\frac{1}{(x-1)^2}}{h}

Getting rid of the fractions within fractions by multiplying both by the highest common denominator:

\frac{(x-1)^2-(x+h-1)^2}{h(x-1)^2(x+h-1)^2}

Did you understand how to get to this point? It's really easy from here, just expand the numerator and simplify.
 


expanding the numerator ,,,

how to multiply h(x^2-2x+1)(x^2+h^2+1)

start with which one ??
 


No no the numerator is the top part of the fractions, if you plug h=0 into the bottom, that h factor makes it all zero, but getting rid of that it won't be zero for h=0 anymore which means we must be able to cancel a factor of h from the numerator.
 


ohh,,,

so i will get -2x-h^2/h(x-1)^2 (x+h-1)^2 ,,,, yet there is "-2x- " so i can't take h with h ,,,

so should i divide all by -2x ?
 


Show me the steps on how you expanded the numerator please, then I can know where you went wrong.
 


it's (h+x-1)^2 -(x-1)^2 ====> h^2 -2xh +2xh +x^2 -1 - ( x^2 -2x+1)2xh will go with -2xh so it's going to be h^2 +2x
 


I don't know what your thought process was, because it's far from being right.
You can try expand this in a few different ways:

You already know that (a-b)^2=a^2-2ab+b^2 so if we let a=x+h and b=1 then (x+h-1)^2=(x+h)^2-2(x+h)+1

or

Expansions work as follows, given two factors (a_1+a_2+...+a_n)(b_1+b_2+...+b_n)
to expand them we simply do it in parts as such,
a_1(b_1+...+b_n)+a_2(b_1+...+b_n)+...+a_n(b_1+...+b_n)

So for (x+h-1)^2 we look at this as (x+h-1)(x+h-1) and then expand by multiplying the 1st term in the 1st factor by everything in the 2nd factor, the 2nd term in the 1st factor by everything in the 2nd factor... etc.

x(x+h-1)+h(x+h-1)-1(x+h-1)
 


Mentallic said:
I don't know what your thought process was, because it's far from being right.
You can try expand this in a few different ways:

You already know that (a-b)^2=a^2-2ab+b^2 so if we let a=x+h and b=1 then (x+h-1)^2=(x+h)^2-2(x+h)+1

or

i know that ,,, but i expand (x+h)^2 also ,,,

that why i got " h^2 -2xh +2xh +x^2 -1 - ( x^2 -2x+1) "
 
  • #10


i saw my mistake at last !
 
  • #11


Hah :smile:

Ok just one last thing, previously you said it's (h+x-1)^2 -(x-1)^2, but it's actually (x-1)^2 - (h+x-1)^2
 

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