# Derivative of 1/x the old fashioned way

• 1MileCrash
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.
1MileCrash

## 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??

You must calculate

$$\lim_{h\rightarrow 0}{\frac{\frac{1}{x+h}-\frac{1}{x}}{h}}$$

First simplify it (to the point where there's only one fraction, instead of three fractions).

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.

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)

1MileCrash said:
That gets me no where.

Yes it will

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

or

-h / (x^2+hx)

And now you still need to divide through h...

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.

That's where I'm stuck.

How can I get rid of the complex fraction here?

"Dividing through h" doesn't seem possible.

$$\lim_{h\rightarrow 0}{\frac{\frac{1}{x+h}-\frac{1}{x}}{h}} = \lim_{h \to 0} \frac{-h}{x^2 + hx}\cdot \frac{1}{h}$$

Can you simplify a bit more and then take the limit?

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

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!

## 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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