Damn differentiation

Hey there,

Could someone try to differentiate

(1 + x^3)^(1/2)
----------------
x^2

I get an answer but it needs simplifying and with god as my witness I cannot get it.

Its mean to be the square root of (1 + x^3) so i just put to the power of 1/2.

Thanks in advance

http://www.vbforums.com/attachment.php?s=&postid=792109

Hope this helps.

Hey there,

The answer in the book gives

-(x^3 + 4)
--------------
2x^3(1+x^3)^1/2

Dont know if you can convert it into that or not?

Hmm... I don't know. Possibly. ;)

Are you sure that is the correct answer? In its SIMPLEST form? I don't know how you could arrive at the answer given by the book... Anyone got any inputs? Am i missing something?

Helger thats how i have been doing it using the product and quotient rule but you dont get the answer. Thanks for the help both of you anyhow, still trying to find the answer though.

the one is the quotient rule alright. However the other is NOT the product rule. Repeat it is NOT. It is the [Kettenregel] (don't know the English term. Maybe 'chain rule'?)

It is a function of a function. NOT a product of two functions. Maybe have another look.

Helger

I know, its the chain rule:o

Just not thinking about what im writing, it still doesnt give whats in the book. Try it and see.

glad you didn't just mix those up. things like that suck most.

My result is different from the book too :(

I got (4x^3 - 2) / (x^3 * (1+x^3)^0.5)



Helger

Ill ask Bill (My maths teacher tomorrow) hell show me, he set it as an optional homework question and he does them all himself to check so hell know how to do it, it was listed as one of the harder questions though. Ill try again at it later.

Please post the result. I'm curious now myself.

Helger

let y = [(1 + x³)^(1/2)] / x²

then y*x² = [(1 + x³)^(1/2)]

Squaring both sides,

y² * x⁴ = 1 + x³

Diff.with respect to x

[2y*(dy/dx)*x⁴] + [y² * (4x³)] = 3x²

2y*(dy/dx)*x⁴ = 3x² - 4(x³)(y²)

dy/dx = [3*x² - 4(x³)(y²)] / [2*x⁴*y]

// since y = [(1 + x³)^(1/2)] / x²

// and y² = (1 + x³) / x⁴

dy/dx = [3*x² - ( 4(x³)*(1+x³)/x⁴ )] / [2*x⁴ * (1 + x³)^(1/2) / x²]

dy/dx = [3*x² - ( 4(1 + x³)/x )] / [ 2*x²*(1+x³)^(1/2)]

dy/dx = [3*x³ - 4 - 4*x³]/[2*x³*(1+x³)^(1/2)]

dy/dx = -[x³ + 4]/[2x³ * (1+x³)^(1/2)]

Hey there,

Although that does give the right answer, we hadnt learned implicity differentiation until this week and so I have kept at it and finally got it.

Heres my working although its a bit sloppy.

Its a bit big to include in the post.