Here we present the proof of the following theorem:

*Let be functions that are differentiable at some . If , then is differentiable at with*

*Quotient Rule*

**Remark:** In the Leibniz notation,

**Proof:** Let :

Letting on both sides:

J.P. McCarthy: Math Page

Last year's maths is easy, this year's maths is hard and next year's maths is impossible.

Here we present the proof of the following theorem:

*Let be functions that are differentiable at some . If , then is differentiable at with*

*Quotient Rule*

**Remark:** In the Leibniz notation,

**Proof:** Let :

Letting on both sides:

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## 5 comments

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May 30, 2011 at 11:55 am

jim kellyThere is a far easier way to do the quotient rule using delta x

May 30, 2011 at 12:30 pm

J.P. McCarthyJim,

Far easier?? Care to elaborate? Surely the difference is just cosmetic?

J.P.

May 30, 2011 at 6:29 pm

J.P. McCarthyJim,

I presume this is it:

http://www.rte.ie/exams/mathsnotes/easyway.pdf

I have to say I like this method — moreover it works for the sum rule also.

Thank you,

J.P.

August 29, 2013 at 7:15 pm

Student 46Hi J.P,

I used the quotient rule to find to find the critical points and this gave me

What do I need to do next?

August 29, 2013 at 7:23 pm

J.P. McCarthyCritical points are points such that so you want

.

Now the only time that a fraction is equal to zero is when the top (numerator) is zero.

Therefore we want

It is difficult to see when a sum is zero but easy to see when a product is zero by the No-Zero-Divisors-Theorem which says that if and are two numbers and then either or . So we want to write the above as a product… we must factorise. What is common to both… the and :

or

or

or

These are the critical points. They might be maxima or minima (or neither — won’t happen on your paper). To do this we test them at the second derivative which is the derivative of the first derivative:

.

Now evaluate and .

If is a critical point then

implies that there is a local minimum at .

implies that there is a local maximum at .

Regards,

J.P.