Differentiating inverse functions (1 Viewer)

[Premus]

Hey

How do you differentiate:

2 sin^-1 (x)^0.5 - sin^-1 (2x -1 )


and show that the derivative = 0
for 0 < x <1

Thanks!!

 

[mojako]

[Attention: this post doesn't give you any working solution... you're free to read but don't expect much.. I post it anyway because I've wasted a few minutes on this and some techniques here may be useful for a few readers, though maybe not for the majority of them]

-------------------
umm do you mean to differentiate
2*sqrt(inverse sine of x) - inverse sine of (2x-1)??

let's differentiate sqrt(inverse sine of x) first, which is (inverse sine of x)^0.5.
Remember when you differentiate things like (x^2+1)^0.5, what you do is you take 1 away from the 0.5 power etc etc like this:
0.5 * (x^2+1)^(-0.5) * 2x
Similarly, the derivative of (inverse sine of x)^0.5 is
0.5 * (inverse sine of x)^(-0.5) * 1/sqrt(1-x^2)

So, the derivative of 2*sqrt(inverse sine of x) is 2 times that thing we get in the previous paragraph.

Now let's differentiate inverse sine of (2x-1). Use the chain rule if you're familiar with that. If not then you probably recognise my following lines anyway.. well it's basically the chain rule but without spelling out the dy/du and du/dx "junks" (well they're not junks... they may prove to be useful for more compex functions)...
1/sqrt[1-(2x-1)^2] * derivative of (2x-1)
= 2/sqrt[1-(2x-1)^2]

Remember the original function is "- inverse sine of (2x-1)", so let's put a minus sign before the thing we get in the previous paragraph.

Now the derivative of the whole function is
1 / [sqrt(1-x^2) * (inverse sine of x)^0.5] - 2/sqrt[1-(2x-1)^2]

Oops... I don't think it equals zero for 0 < x < 1...
maybe I misunderstood your question or made a mistake in the calculation.

---------------------------------------
Mmm I think you mean inverse sine of x^0.5... let's see...
its derivative is
1/sqrt[1-(x^0.5)^2] * derivative of x^0.5
= 1/sqrt(1-x^2) * 0.5x^(-0.5)

Well I give up.. Sorry..

 

[:: ck ::]

nothing wrong with the question

here you go :

 

[Premus]

Hi

I dont understand how u went from

d / dx ( 2 sin-1 root x) = 2 * 1 / 2root x * 1 / root (1-x)

---------------------------------

I thought its just 2 * 1 / root (1-x)

Thanks

 

[mojako]

Read the last part of my previous post...:
>> Mmm I think you mean inverse sine of x^0.5... let's see...
its derivative is
1/sqrt[1-(x^0.5)^2] * derivative of x^0.5
= 1/sqrt(1-x^2) * 0.5x^(-0.5) <<

BTW, you can write inverse sin as "asin", inverse cos as "acos" etc, which is the way you do it in many computer calculators. Not that I guarantee other readers to understand though

 

[:: ck ::]

Hi

I dont understand how u went from

d / dx ( 2 sin-1 root x) = 2 * 1 / 2root x * 1 / root (1-x)

---------------------------------

I thought its just 2 * 1 / root (1-x)

Thanks

i told u in the post.. the squared part didnt come up in my scan....

 

[CM_Tutor]

NOTE : line 2... the squared got stuffed up during scanning for some reason..

Why should there have been a squared in that line? Don't you have 1 / sqrt(1 - sqrt(x)²) = 1 / sqrt(1 - x) at that point?

 

[CM_Tutor]

I dont understand how u went from

d / dx ( 2 sin-1 root x) = 2 * 1 / 2root x * 1 / root (1-x)

---------------------------------

I thought its just 2 * 1 / root (1-x)

You are using the Chain Rule, and have forgotten to differentiate the function inside (ie. the sqrt(x)), which gives the extra 1 / (2 * sqrt(x)) term.

 

[:: ck ::]

hahaha oops XD !!

when Premus said something about my 2nd line being wrong... i thought it didnt pop up in the scan.......................... =.="

silly me ..