Solve derivative by using the definition of the derivative limit

In summary, the student was trying to solve a derivative by using the definition of the derivative limit, but ultimately made a mistake. They were able to correct the mistake by expanding the numerator and simplifying.
  • #1
MrNeWBiE
75
0
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


---------------------------------------------------------------------------

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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  • #2


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,

[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.
 
  • #3


expanding the numerator ,,,

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

start with which one ??
 
  • #4


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.
 
  • #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 ?
 
  • #6


Show me the steps on how you expanded the numerator please, then I can know where you went wrong.
 
  • #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 going to be h^2 +2x
 
  • #8


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]
 
  • #9


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 [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

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
 

Related to Solve derivative by using the definition of the derivative limit

1. What is the definition of the derivative limit?

The derivative limit is the instantaneous rate of change of a function at a specific point. It is found by taking the limit of the difference quotient as the change in the independent variable approaches zero.

2. How is the derivative limit used to solve derivatives?

The derivative limit is used to find the slope of a tangent line at a specific point on a curve. This slope is equal to the value of the derivative at that point. By using the definition of the derivative limit, we can find the exact value of the derivative at any point on a curve.

3. What is the difference between the derivative limit and the derivative?

The derivative limit is the concept of finding the slope of a tangent line at a specific point on a curve. The derivative is a function that gives us the slope of the tangent line at any point on the curve. In other words, the derivative limit is the process of finding the derivative at a specific point.

4. Why is the definition of the derivative limit important?

The definition of the derivative limit is important because it is the foundation of calculus and allows us to find the instantaneous rate of change of a function. It also helps us understand the behavior of a function at a specific point and make predictions about its values in the future.

5. Can the derivative limit be used to find the derivative of any function?

Yes, the derivative limit can be used to find the derivative of any function, as long as the limit exists. However, for more complex functions, other methods such as the power rule or product rule may be more efficient to use. The derivative limit is most useful for finding the derivative of functions with non-constant rates of change.

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