Derivative of 1/x the old fashioned way

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In summary: So as h approaches 0, hx approaches zero, leaving me with x^2 in the denomenator for a final result of 1/x^2.
  • #1
1MileCrash
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Homework Statement



f(x) = 1/x

Find the derivative

Homework Equations



f'(x)= lim h->0

[f(x+h) - f(x)] / [h]

The Attempt at a Solution



I MUST show work and use that formula to arrive at the derivative (1/x^2)

I can't figure out the algebra.

I rearrange it countless ways..

[1/(x+h) - 1/x] - (1/x)] / [h]

ETC.

But I can NEVER find a way to algebraically arrive at the derivative! Ever!

What would you guys do to algebraically solve this??
 
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  • #2
You must calculate

[tex]\lim_{h\rightarrow 0}{\frac{\frac{1}{x+h}-\frac{1}{x}}{h}}[/tex]

First simplify it (to the point where there's only one fraction, instead of three fractions).
 
  • #3
Rearranging your limit expression is not necessarily going to help unless you can do the subtraction [1/(x+h) - 1/x] correctly. Show what you get for this calculation.
 
  • #4
SteamKing said:
Rearranging your limit expression is not necessarily going to help unless you can do the subtraction [1/(x+h) - 1/x] correctly. Show what you get for this calculation.

That gets me no where.

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

or

-h / (x^2+hx)
 
  • #5
1MileCrash said:
That gets me no where.

Yes it will :smile:

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

or

-h / (x^2+hx)

And now you still need to divide through h...
 
  • #6
1MileCrash said:
That gets me no where.

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

or

-h / (x^2+hx)
This is not the complete quotient. This is just the simplification of the numerator, 1/x - 1/(x + h)

micromass said:
And now you still need to divide through h...
Dividing by h is equivalent to multiplying by 1/h.
 
  • #7
That's where I'm stuck.

How can I get rid of the complex fraction here?

"Dividing through h" doesn't seem possible.
 
  • #8
[tex]\lim_{h\rightarrow 0}{\frac{\frac{1}{x+h}-\frac{1}{x}}{h}} = \lim_{h \to 0} \frac{-h}{x^2 + hx}\cdot \frac{1}{h}[/tex]

Can you simplify a bit more and then take the limit?
 
  • #9
NOW I see! Thanks!

-h / hx^2 + h^2x

=

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

(-1) / x^2 + hx

as h approaches 0, hx approaches zero, leaving me with x^2 in the denomenator for a final result of

- 1/x^2
 
  • #10
1MileCrash said:
NOW I see! Thanks!

-h / hx^2 + h^2x

=

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

(-1) / x^2 + hx

as h approaches 0, hx approaches zero, leaving me with x^2 in the denomenator for a final result of

- 1/x^2

Right!
 

FAQ: Derivative of 1/x the old fashioned way

What is the derivative of 1/x the old fashioned way?

The derivative of 1/x the old fashioned way is -1/x^2.

What is the "old fashioned way" of finding the derivative of 1/x?

The "old fashioned way" of finding the derivative of 1/x is by using the quotient rule, which states that the derivative of f(x)/g(x) is (g(x)f'(x) - f(x)g'(x)) / g(x)^2.

Why is it important to know the derivative of 1/x?

Knowing the derivative of 1/x is important in many mathematical and scientific applications, such as in physics, engineering, and economics. It allows us to understand how a variable changes in relation to another variable, and helps us to solve problems involving rates of change.

Is there a simpler way to find the derivative of 1/x?

Yes, there is a simpler way to find the derivative of 1/x. By using the power rule, we can rewrite 1/x as x^-1 and then take the derivative, which results in -1*x^-2 or -1/x^2. This method is often preferred over the quotient rule for its simplicity.

Can the "old fashioned way" of finding the derivative of 1/x be applied to other functions?

Yes, the quotient rule can be applied to any function that can be expressed as a quotient of two other functions. However, for simpler functions, like 1/x, it is often more efficient to use the power rule.

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