| Thread Closed |
solve derivative by using the definition of the derivative limit ,,, |
Share Thread | Thread Tools |
| Oct12-10, 01:55 AM | #1 |
|
|
solve derivative by using the definition of the derivative limit ,,,
hi alll
1. The problem statement, all variables and given/known data[/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 --------------------------------------------------------------------------- can someone tell me what did i do wrong so i can't devide the H down and the one up ,,, i been trying to solve it from 2days ,,, =,= |
| Oct12-10, 02:07 AM | #2 |
|
Recognitions:
|
Uhh... let's ignore where the mistake was (the hell I'm gonna try read through that mess) and will simply guide you through it instead.
So this is what we have, [tex]\frac{\frac{1}{(x+h-1)^2}-\frac{1}{(x-1)^2}}{h}[/tex] Getting rid of the fractions within fractions by multiplying both by the highest common denominator: [tex]\frac{(x-1)^2-(x+h-1)^2}{h(x-1)^2(x+h-1)^2}[/tex] Did you understand how to get to this point? It's really easy from here, just expand the numerator and simplify. |
| Oct12-10, 02:17 AM | #3 |
|
|
expanding the numerator ,,,
how to multiply h(x^2-2x+1)(x^2+h^2+1) start with which one ?? |
| Oct12-10, 03:08 AM | #4 |
|
Recognitions:
|
solve derivative by using the definition of the derivative limit ,,,
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.
|
| Oct12-10, 07:43 AM | #5 |
|
|
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 ? |
| Oct12-10, 08:06 AM | #6 |
|
Recognitions:
|
Show me the steps on how you expanded the numerator please, then I can know where you went wrong.
|
| Oct12-10, 08:19 AM | #7 |
|
|
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 gonna be h^2 +2x |
| Oct12-10, 08:47 AM | #8 |
|
Recognitions:
|
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 [tex](a-b)^2=a^2-2ab+b^2[/tex] so if we let a=x+h and b=1 then [tex](x+h-1)^2=(x+h)^2-2(x+h)+1[/tex] or Expansions work as follows, given two factors [tex](a_1+a_2+...+a_n)(b_1+b_2+...+b_n)[/tex] to expand them we simply do it in parts as such, [tex]a_1(b_1+...+b_n)+a_2(b_1+...+b_n)+...+a_n(b_1+...+b_n)[/tex] So for [tex](x+h-1)^2[/tex] we look at this as [tex](x+h-1)(x+h-1)[/tex] 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. [tex]x(x+h-1)+h(x+h-1)-1(x+h-1)[/tex] |
| Oct12-10, 10:12 AM | #9 |
|
|
that why i got " h^2 -2xh +2xh +x^2 -1 - ( x^2 -2x+1) " |
| Oct12-10, 10:17 AM | #10 |
|
|
i saw my mistake at last !!
|
| Oct12-10, 10:21 AM | #11 |
|
Recognitions:
|
Hah
![]() 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 |
| Thread Closed |
| Thread Tools | |
Similar Threads for: solve derivative by using the definition of the derivative limit ,,,
|
||||
| Thread | Forum | Replies | ||
| Limit Definition of Derivative | Calculus & Beyond Homework | 1 | ||
| Multivariable Limit (Definition of Derivative) | Calculus & Beyond Homework | 1 | ||
| Definition of a Derivative/Limit | Calculus & Beyond Homework | 0 | ||
| Derivative of 10^x using limit definition | Calculus & Beyond Homework | 16 | ||
| Using limit definition of multivariable derivative | Calculus | 2 | ||