PLEASE HELP ME TO SOLVE THIS QUESTION

[r267747]

-3/4
x
I don't know how to find its derivative using first principle. So please help me

 

[galactus]

What is that supposed to be?. Please use preview before you post so you can see how it will turn out.

 

[r267747]

I don't know how to find its derivative using first principle with respect to x. So please help me
-3/4
x (where-3/4 is the power of x)

 

[soroban]

Hello, r267747!

This is an ugly problem!

Find the derivative using first principles: .$\displaystyle f(x) \:=\:x^{-\frac{3}{4}}$

\(\displaystyle \text}{We want: }\;f'(x) \:=\:\lim_{_{h\to0}}\dfrac{f(x+h) - f(x)}{h}\)

$\displaystyle [1]\;\;f(x+h) - f(x) \;=\;(x+h)^{-\frac{3}{4}} - x^{-\frac{3}{4}} \;=\;\frac{1}{(x+h)^{\frac{3}{4}}} - \frac{1}{x^{\frac{3}{4}}} \;=\;\frac{x^{\frac{3}{4}} - (x+h)^{\frac{3}{4}}}{x^{\frac{3}{4}}(x+h)^{\frac{3}{4}}}$


Here's where it gets very messy . . .


$\displaystyle \text{The numerator is: }\:N \;=\;x^{\frac{3}{4}} - (x+h)^{\frac{3}{4}}$

$\displaystyle \text{Let: }\:a \,=\,x^{\frac{1}{4}},\;b \,=\,(x+h)^{\frac{1}{4}}$

. . $\displaystyle \text{and we have: }\;N \;=\;a^3 - b^3$

. . $\displaystyle \text{which factors: }\;N \;=\;(a-b)(a^2+ab+b^2)$


$\displaystyle \text{The fraction becomes: }\;\frac{(a-b)(a^2+ab+b^2)}{a^3b^3}$


$\displaystyle \text{Multiply by: }\:\frac{a^3 + a^2b+ab^2+b^3}{a^3+a^2b+ab^2+b^3}$

. . $\displaystyle \frac{(a-b)(a^2+ab+b^2)}{a^3b^3}\cdot \frac{a^3 + a^2b+ab^2+b^3}{a^3+a^2b+ab^2+b^3}$

. . . . .$\displaystyle =\;\frac{(a-b)(a^3+a^2b+ab^2+b^3)}{a^3b^3}\cdot\frac{a^2+ab+b^2}{a^3+a^2b+ab^2+b^3}$

. . . . .$\displaystyle =\;\frac{a^4-b^4}{a^3b^3}\cdot\frac{a^2+ab+b^2}{a^3+a^2b+ab^2+b^3}$


$\displaystyle \text{Back-substitute in the first fraction:}$

. . $\displaystyle \frac{\left[x^{\frac{1}{4}}\right]^4 - \left[(x+h)^{\frac{1}{4}}\right]^4}{a^3b^3}\cdot\frac{a^2+ab + b^2}{a^3+a^2b+ab^2+b^3}$

. . $\displaystyle =\;\frac{x-(x+h)}{a^3b^3}\cdot\frac{a^2+ab+b^2}{a^3+a^2b + ab^2 + b^3} \;=\;\frac{-h}{a^3b^3}\cdot\frac{a^2+ab+b^2}{a^3+a^2b+ab^2+b^3}$


$\displaystyle [2]\;\text{ Divide by }h\!:\quad \frac{f(x+h)-f(x)}{h} \;=\;\frac{-(a^2+ab+b^2)}{a^3b^3(a^3+a^2b+ab^2+b^3)}$


$\displaystyle \text{Back-substitute: }\; -\frac{ x^{\frac{1}{2}} + (x^{\frac{1}{4}})(x+h)^{\frac{1}{4}} + (x+h)^{\frac{1}{2}}} {x^{\frac{3}{4}}(x+h)^{\frac{3}{4}}\left[x^{\frac{3}{4}} + x^\frac{1}{2}(x+h)^{\frac{1}{4}} + x^{\frac{1}{4}}(x+h)^{\frac{1}{2}} + (x+h)^{\frac{3}{4}}\right]}$


$\displaystyle [3]\;\text{ Take limit: }\;\lim_{h\to0}\left( -\,\frac{ x^{\frac{1}{2}} + (x^{\frac{1}{4}})(x+h)^{\frac{1}{4}} + (x+h)^{\frac{1}{2}}} {x^{\frac{3}{4}}(x+h)^{\frac{3}{4}}\left[x^{\frac{3}{4}} + x^\frac{1}{2}(x+h)^{\frac{1}{4}} + x^{\frac{1}{4}}(x+h)^{\frac{1}{2}} + (x+h)^{\frac{3}{4}}\right]} \right)$

$\displaystyle \text{And we have: }\;-\, \frac{x^{\frac{1}{2}} + (x^{\frac{1}{4}})(x^{\frac{1}{4}}) + x^{\frac{1}{2}}} {(x^{\frac{3}{4}}) (x^{\frac{3}{4}}) \left(x^{\frac{3}{4}} + (x^{\frac{1}{2}})(x^{\frac{1}{4}}) + (x^{\frac{1}{4}})(x^{\frac{1}{2}}) + x^{\frac{3}{4}}\right)} \;=\;-\,\frac{x^{\frac{1}{2}} + x^{\frac{1}{2}} + x^{\frac{1}{2}}} {x^{\frac{3}{2}}\left(x^{\frac{3}{4}} + x^{\frac{3}{4}} + x^{\frac{3}{4}} + x^{\frac{3}{4}}\right)}$

. . . . . . . . $\displaystyle =\;-\,\frac{3x^{\frac{1}{2}}} {x^{\frac{3}{2}}\cdot4x^{\frac{3}{4}}} \;=\;-\,\frac{3x^{\frac{1}{2}}} {4x^{\frac{9}{4}}}$


$\displaystyle \text{Therefore: }\;f'(x) \;=\;\frac{3}{4x^{\frac{7}{4}}} \;=\;\frac{3}{4}x^{-\frac{7}{4}}$

 

[galactus]

Fantastic Soroban. That was a rough one for a Calc I student.

It's a shame L'Hopital couldn't be used.

$\displaystyle \frac{(x+h)^{\frac{-3}{4}}-x^{\frac{-3}{4}}}{h}$

$\displaystyle \lim_{h\to 0}\frac{-3}{4}(x+h)^{\frac{-7}{4}}$

$\displaystyle \frac{-3}{4}x^{\frac{-7}{4}}$

Just being wacky. Rather circular logic, huh

 

[BigGlenntheHeavy]

$\displaystyle Another \ way, \ I \ think \ a \ little \ less \ protracted.$

$\displaystyle f(x) \ = \ x^{-3/4}, \ find \ f'(x) \ the \ hard \ way.$

$\displaystyle f'(x) \ = \ \lim_{h\to0}\frac{f(x+h)-f(x)}{h} \ = \ \lim_{h\to0}\frac{(x+h)^{-3/4}-x^{-3/4}}{h}$

$\displaystyle After \ some \ algebra, \ we \ get: \ \lim_{h\to0} \ \frac{x^{3/4}-(x+h)^{3/4}}{h(x^{3/4})(x+h)^{3/4}}$

$\displaystyle Now, \ a \ minor \ digression.$

$\displaystyle Note, \ now \ here \ is \ the \ kicker: \ (a-b) \ = \ (a^{1/4}-b^{1/4})(a^{3/4}+a^{1/4}b^{1/2}+a^{1/2}b^{1/4}+b^{3/4})$

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

$\displaystyle Let \ a^{1/4} \ = \ x^{3/4} \ and \ b^{1/4} \ = \ (x+h)^{3/4}$

$\displaystyle Then, \ a^{1/2} \ = \ x^{3/2} \ and \ b^{1/2} \ = \ (x+h)^{3/2}$

$\displaystyle And \ a^{3/4} \ = \ x^{9/4} \ and \ b^{3/4} \ = \ (x+h)^{9/4}$

$\displaystyle Finally, \ a \ = \ x^{3} \ and \ b \ = \ (x+h)^{3}, \ digression \ over.$

$\displaystyle Ergo, \ \lim_{h\to0} \ \frac{x^{3/4}-(x+h)^{3/4}}{h(x^{3/4})(x+h)^{3/4}} \ = \ \lim_{h\to0} \ \frac{a^{1/4}-b^{1/4}}{h(a^{1/4})(b^{1/4})}, \ Substitute$

$\displaystyle = \ \lim_{h\to0} \ \frac{(a-b)}{h[a^{3/4}+a^{1/4}b^{1/2}+a^{1/2}b^{1/4}+b^{3/4}][a^{1/4}b^{1/4}]}, \ Substitute \ (Line \ 7)$

$\displaystyle = \ \lim_{h\to0} \ \frac{(a-b)}{h[ab^{1/4}+a^{1/2}b^{3/4}+a^{3/4}b^{1/2}+a^{1/4}b]}$

$\displaystyle = \ \lim_{h\to0} \ \frac{x^{3}-(x+h)^{3}}{h[x^{3}(x+h)^{3/4}+x^{3/2}(x+h)^{9/4}+x^{9/4}(x+h)^{3/2}+x^{3/4}(x+h)^{3}]}, \ Resubstitute$

$\displaystyle = \ \lim_{h\to0} \ \frac{x^{3}-(x^{3}+3x^{2}h+3xh^{2}+h^{3})}{h[x^{3}(x+h)^{3/4}+x^{3/2}(x+h)^{9/4}+x^{9/4}(x+h)^{3/2}+x^{3/4}(x+h)^{3}]}, \ Expand \ binomial$

$\displaystyle = \ \lim_{h\to0} \ \frac{-3x^{2}-3xh-h^{2}}{[x^{3}(x+h)^{3/4}+x^{3/2}(x+h)^{9/4}+x^{9/4}(x+h)^{3/2}+x^{3/4}(x+h)^{3}]}, \ Cancel \ out \ h$

$\displaystyle = \ \lim_{h\to0} \ \frac{-3x^{2}}{(x^{3})(x^{3/4})+(x^{3/2})(x^{9/4})+(x^{9/4})(x^{3/2})+(x^{3/4})(x^{3})}, \ Self-evident$

$\displaystyle = \ \lim_{h\to0} \ -\frac{3x^{2}}{x^{15/4}+x^{15/4}+x^{15/4}+x^{15/4}} \ = \ \lim_{h\to0} \ -\frac{3x^{2}}{4x^{15/4}} \ = \ -\frac{3}{4}x^{-7/4} \ QED$

 

[BigGlenntheHeavy]

$\displaystyle An \ easier \ way \ yet. \ The \ easiest?$

$\displaystyle From \ above, \ we \ have \ f'(x) \ = \ \lim_{h\to0} \ \frac{x^{3/4}-(x+h)^{3/4}}{h(x^{3/4})(x+h)^{3/4}}$

$\displaystyle = \ \lim_{h\to0} \ \frac{x^{3/4}-(x+h)^{3/4}}{h(x^{3/4})(x+h)^{3/4}} \ * \ \frac{x^{3/4}+(x+h)^{3/4}}{x^{3/4}+(x+h)^{3/4}} \ = \ \lim_{h\to0} \ \frac{x^{3/2}-(x+h)^{3/2}}{h[x^{3/2}(x+h)^{3/4}+x^{3/4}(x+h)^{3/2}]}$

$\displaystyle = \ \lim_{h\to0} \ \frac{x^{3/2}-(x+h)^{3/2}}{h[x^{3/2}(x+h)^{3/4}+x^{3/4}(x+h)^{3/2}]} \ * \ \frac{x^{3/2}+(x+h)^{3/2}}{x^{3/2}+(x+h)^{3/2}}$

$\displaystyle = \ \lim_{h\to0} \ \frac{x^{3}-(x+h)^{3}}{h[x^{3/2}(x+h)^{3/4}+x^{3/4}(x+h)^{3/2}][x^{3/2}+(x+h)^{3/2}]}$

$\displaystyle = \ \lim_{h\to0} \ \frac{x^{3}-x^{3}-3x^{2}h-3xh^{2}-h^{3}}{h[x^{3/2}(x+h)^{3/4}+x^{3/4}(x+h)^{3/2}][x^{3/2}+(x+h)^{3/2}]}$

$\displaystyle = \ \lim_{h\to0} \ \frac{-3x^{2}-3xh-h^{2}}{[x^{3/2}(x+h)^{3/4}+x^{3/4}(x+h)^{3/2}][x^{3/2}+(x+h)^{3/2}]}$

$\displaystyle = \ \lim_{h\to0} \ \frac{-3x^{2}}{[x^{3/2}x^{3/4}+x^{3/4}x^{3/2}][x^{3/2}+x^{3/2}]} \ = \ \lim_{h\to0} \ \frac{-3x^{2}}{(2x^{9/4})(2x^{3/2})} \ = \ \lim_{h\to0} \ \frac{-3x^{2}}{4x^{15/4}} \ =-\frac{3}{4}x^{-7/4}QED$

$\displaystyle Note: \ f'(x) \ = \ \lim_{h\to0} \ \frac{x^{3/4}-(x+h)^{3/4}}{h(x^{3/4})(x+h)^{3/4}} \ \ The \ Marqui \ gives \ the \ indeterminate \ form \ \frac{0}{0},$

$\displaystyle \ but \ to \ no \ avail, \ as \ it \ is \ circular, \ sorry \ galactus.$

 

[galactus]

Nice, Glenn

 

[BigGlenntheHeavy]

$\displaystyle Note: \ If \ I \ plugged \ \lim_{h\to0} \ \frac{x^{3/4}-(x+h)^{3/4}}{h(x^{3/4})(x+h)^{3/4}} \ into \ my \ trusty \ TI-89,$

$\displaystyle \ it \ will \ immediately \ return \ the \ correct \ limit \ - \ \frac{-3}{4x^{7/4}}.$

$\displaystyle Anyone \ know \ how \ it \ does \ that? \ In \ other \ words, \ does \ it \ follow \ my \ above \ "easier" \ limit?$

$\displaystyle I \ would \ like \ to \ see \ the \ algorithm \ it \ follows. \ Just \ curious.$

 

[BigGlenntheHeavy]

The easier way possible to find the limit of:

$\displaystyle \lim_{h\to0} \ \frac{(x+h)^{-3/4}-x^{-3/4}}{h}. \ Just \ plugged \ it \ into \ your \ trusty \ TI-89$

\(\displaystyle and \ avoid \ all \ the \ grunt \ work. \ It \ (TI-89) \ will \ immediately \ spit \ out \ the \ answer, \ to \ wit: \ \frac{-3}{4x^{7/4}}.\\)

$\displaystyle Afterthought: \ This \ is \ what \ drove \ "The \ Unabomber" \ to \ distraction. \ He \ could \ not \ take \ the$

$\displaystyle marvels \ of \ modern \ technology \ (he \ had \ a \ Doctorate \ in \ Math), \ and \ hated \ anything \ to \ do \ with$

$\displaystyle computers \ as \ he \ most \ likely \ grunted \ everything \ out \ and \ expected \ the \ same \ from \ his$

$\displaystyle contemporaries.$

$\displaystyle Before \ being \ captured, \ he \ lived \ in \ a \ shack \ in \ the \ woods, \ and \ ate \ beavers \ for \ his \ sustenance. \ Ugh.$

 

[galactus]

Yes, but sometimes it's fun to do it by hand to keep them synapses firing. I like my calculators, but I do not want to become too dependent on them.

 

[BigGlenntheHeavy]

Right on galactus, as I concur wholeheartedly, but one does wonder what is happening to the youth.

Case in point: When I lived in New Orleans, I once went into a 7-11 to purchase some junk food, but was unable to as the cash register (computized) was on the blink and the clerk wasn't adept enough to add up my purchases manually.

A blessing in disguise?

 

[galactus]

The so-called 'youth'. Yeah, I see it all the time. The apathy and downright laziness is very disconcerting.

I know a young man right now that just sits on his behind playing video games instead of trying to better his life. No education, no skills, and then gripes because he can't get a job. "The ones I find suck", he said. No wonder. Who wants someone like that?. As far as a prospective employer is concerned, he's worthless. Going to college or trade school is just crazy talk. Yet, there are plenty of people out there with skills and education who can not find any job.

Take math, for instance. Do you think society as a whole, indirectly or directly, programs people to hate it so?. Todays American way of 'thinking' appears to be stay as dumbed down as possible. The TV programs that get the high ratings are a testament to that. Any exertion, rather physical or mental, is to be avoided at all cost. Let's all just party and play video games for a living. That would be nice, but that's not the way it is.

And another thing is that infernal text messaging that has spilled over into the way people write in general.

My use of grammar is not the best, that's for sure, but :roll:

 

[BigGlenntheHeavy]

Galactus, enough said, as I know exactly where you are coming from.

Think of "The Time Machine" by H.G. Wells. Remember, all the people lived in a "state of ignorant bliss" controlled by the Murdocks, if memory serves me correctly.

 

[galactus]

Very good analogy.

It was the Morlocks.