Another one

\(\displaystyle Find \ the \ derivative \ using \ first \ principles: \ f(x) \ = \ x^{-3/5}.\)

\(\displaystyle A \ good \ exercise \ in \ the \ manipulation \ of \ algebraic \ symbols, \ unless \ you \ know \ an \ easier \ method.\)

\(\displaystyle galactus, \ this \ is \ how \ I \ approach \ this \ problem. \ As \ always \ I \ am \ open \ to \ constructive\)

\(\displaystyle criticism \ from \ whom \ ever; \ basically \ is \ there \ another, \ easier \ way?\)

\(\displaystyle Note: \ My \ way \ is \ the \ same \ as \ yours, \ except \ for \ minor \ differences, \ to \ wit:\)

\(\displaystyle f(x) \ = \ x^{-3/5}, \ find \ the \ derivative \ using \ first \ principle.\)

\(\displaystyle f'(x) \ = \ \lim_{h\to0} \ \frac{(x+h)^{-3/5}-x^{-3/5}}{h} \ which \ gives \ \lim_{h\to0} \ \frac{x^{3/5}-(x+h)^{3/5}}{hx^{3/5}(x+h)^{3/5}}\)

\(\displaystyle Now \ (a-b) \ = \ (a^{1/5}-b^{1/5})(a^{4/5}+a^{1/5}b^{3/5}+a^{2/5}b^{2/5}+a^{3/5}b^{1/5}+b^{4/5})\)

\(\displaystyle This \ implies \ that \ (a^{1/5}-b^{1/5}) \ = \ \frac{(a-b)}{(a^{4/5}+a^{1/5}b^{3/5}+a^{2/5}b^{2/5}+a^{3/5}b^{1/5}+b^{4/5})}\)

\(\displaystyle Ergo \ let \ a^{1/5} \ = \ x^{3/5} \ and \ b^{1/5} \ = \ (x+h)^{3/5}\)

\(\displaystyle By \ substitution, \ we \ have \ - \ \lim_{h\to0} \ \frac{a^{1/5}-b^{1/5}}{ha^{1/5}b^{1/5}}\)

\(\displaystyle Substituting \ again \ gives \ - \ \lim_{h\to0} \ \frac{a-b}{h[a^{1/5}b^{1/5}][a^{4/5}+a^{1/5}b^{3/5}+a^{2/5}b^{2/5}+a^{3/5}b^{1/5}+b^{4/5}]}\)

\(\displaystyle = \ \lim_{h\to0} \ \frac{x^{3}-(x+h)^{3}}{h[x^{3/5}(x+h)^{3/5}][x^{12/5}+x^{3/5}(x+h)^{9/5}+x^{6/5}(x+h)^{6/5}+x^{9/5}(x+h)^{3/5}+(x+h)^{12/5}]}\)

\(\displaystyle = \ \lim_{h\to0} \ \frac{x^{3}-x^{3}-3x^{2}h-3xh^{2}-h^{3}}{h[x^{3/5}(x+h)^{3/5}][x^{12/5}+x^{3/5}(x+h)^{9/5}+x^{6/5}(x+h)^{6/5}+x^{9/5}(x+h)^{3/5}+(x+h)^{12/5}]}\)

\(\displaystyle = \ \lim_{h\to0} \ \frac{-3x^{2}-3xh-h^{2}}{[x^{3/5}(x+h)^{3/5}][x^{12/5}+x^{3/5}(x+h)^{9/5}+x^{6/5}(x+h)^{6/5}+x^{9/5}(x+h)^{3/5}+(x+h)^{12/5}]}\)

\(\displaystyle = \ \lim_{h\to0} \ \frac{-3x^{2}}{(x^{3/5})(x^{3/5})[x^{12/5}+(x^{3/5})(x^{9/5})+(x^{6/5})(x^{6/5})+(x^{9/5})(x^{3/5})+x^{12/5}]}\)

\(\displaystyle = \ \lim_{h\to0} \ \frac{-3x^{2}}{x^{6/5}[x^{12/5}+x^{12/5}+x^{12/5}+x^{12/5}+x^{12/5}]} \ = \ \lim_{h\to0} \ \frac{-3x^{2}}{x^{6/5}(5x^{12/5})} \ = \ \lim_{h\to0} \ \frac{-3x^{2}}{5x^{18/5}} \ = \ \frac{-3}{5x^{8/5}}\)

\(\displaystyle Done, \ gasp.\)

soroban, can you think of any easier way, or do these types of problems always have a high grunt factor?

I think the way you done it in your other post is about as easy as any. Where you used the conjugate.

Note, there is a pattern, related to the binomial expansion, depending on what one uses.

i.e. \(\displaystyle a^{5}-b^{5}=(a-b)(a^{4}+a^{3}b+a^{2}b^{2}+ab^{3}+b^{4})\)

Perhaps try to generalize that.

Thanks, galactus, however the conjugate only works, as far as I can see, if the denominator is even.

Post Script: By the way Cody, I wasn't singling you out in my remark about the Marqui, as I was just interjecting a little levity to the board.

I deserved the rebuff for being so vague in my comment about saying L'Hopital didn't work. Poor choice of words.

Did you try the conjugate thing with odd denominator. Maybe I will give something a go in a little while. I just got my 4-wheeler back from getting worked on. I may go for a ride .

\(\displaystyle Cody, \ as \ we \ let \ h=0, \ we \ have \ \frac{0}{0}\)

By jove you're right. I thought I done something stupid, because it worked out so quickly. I left all the h's inside the parenthese go to 0 and then

cancelled the one in the numerator and denominator. DUH. I am deleting that stupidity. It ended up being correct, though

Frankly, I think the way you done it and the way Soroban done it, is as good as any. I always preferred the substitution thing myself, but the algebra on these can be a brute. I am leaving it be. What you and Soroban have done is good enough for me.

EDIT: as a matter of fact, I just went out to the shed where I have my books and papers. I have a calculus book I rooted out that gives an example of these types of problems using first principles. Their method is like what has been posted by you and Soroban.

Yeah, ok, but lets see you try that on \(\displaystyle x^{-\frac{7919}{7907 }}\) :wink:

Is it ok to use a Cray?

daon said:

Yeah, ok, but lets see you try that on \(\displaystyle x^{-\frac{7919}{7907 }}\) :wink:

Click to expand...

Yes, indeed, Daon. But not me

Is it ok to use a Cray?

Click to expand...

What, pratell, is a Cray?. A crayon?. :wink:

The "easy way" is to do the general cases for \(\displaystyle x^m, m \in \mathbb{Z}\) and \(\displaystyle x^{1/n}, n \in \mathbb{N}\). Let \(\displaystyle y=x^{-3}\) and follow suit with \(\displaystyle y^{1/5}\). Or, just prove the chain rule first . Though I think that may not be using "first principles."

And, yeah, what is a Cray?

I was thinking, Cody's suggestion is quite interesting!
Using a Crayon could release massive amounts of childhood creativity!

A Cray is a super computer, I think it is/was used (in tandem) to find fantastic primes and to prove that PI, square root of two, etc. were indeed irrational.

BigGlenntheHeavy said:

A Cray is a super computer, I think it is/was used (in tandem) to find fantastic primes and to prove that PI, square root of two, etc. were indeed irrational.

Click to expand...

Ah, ok. Are there really brute-force type algorithms to prove numbers are irrational? Did computer scientists not believe the mathematicians who've already proved them as such? Interesting.

I have never heard of an algorithm for proving a number is irrational. That would be cool though. Perhaps a supercomputer takes it out to a trillion decimal places, and if it finds no pattern says it's irrational. Who knows. Cool stuff though.

Cody, the Cray didn't "Prove' that the square root of 2 was irrational, it prove up to some fantastic sum that its decimals never ended; which begs the question, is the square root of two irrational or does its decimal representation end at some fantastic rational fraction.

Note: The only proofs that I ever seen that the square root of 2 was irrarional were proofs by contradiction which always left something lacking in my mind; however, just my opinion and you know what they say about opinions.

The idea that they (whoever they are) used the Cray to show that the square root of two "never ends" is food for thought in its own right.

An Indirect proof is still a proof.

Here is an easy proof,

\(\displaystyle x^2=2 \iff x^2-2=0\)

By the Rational Root's Theorem, we can deduce that this equation has 2 solutions which, if rational, must be two of

\(\displaystyle x=\pm 2, \pm 1\)

All of which fail.

Hence the solutions \(\displaystyle x_1,x_2\) of this equation must both not be rational.

Good show daon, you made a believer out of me, the square root of two is indeed irrational.

Why they tried to prove,if they did, otherwise (with the Cray), I have no idea.