Derivative of the arcsin (2x)

[rayroshi]

In learning about how to find the derivative of inverse trig functions, I see two different formulas given, depending upon which text you use: You can, supposedly, use the first formula, d/dx (arcsin u) = u'/(1-u^2)^1/2, or you can use the second formula, d/dx (arcsin u) = 1/(1- x^2)^1/2. (Sorry, but I don't know how to use LaTeX yet)

However, when I try applying them, I get different answers, so I must be doing something wrong. I would appreciate it, if someone would point out my mistake. (By the way, I realize you can use the chain rule to get the answer, but that is an aside and not what I am now trying to figure out at the moment.

Here is an example of what I am talking about:

a) Given Problem: Find the derivative of arcsin of 2x.
b) Using first formula, d/dx arcsin (u) = u'/(1-u^2)^1/2, where u = 2x
= (2x)' / (1- (2x)^2)^1/2
= 2/(1-4x^2)^1/2 This is the correct answer. However, why is this second version, below, wrong?

a) Using the second formula: d/dx arcsin (2x) = 1/(1-x^2)^1/2, where x = 2x
= 1/(1-(2x)^2)^1/2
= 1/(1-4x^2)^1/2 This is wrong, being twice as much as the above, correct answer...but why? I can't see where I have gone wrong, if both of those, given, formulas are correct.

Any help would be much appreciated, as I'm eighty-years-old and am trying to learn calculus by myself, a task that would be nearly impossible, if it weren't for the internet and lots of good books...which isn't really 'by myself,' I guess.

 

[Dr.Peterson]

In learning about how to find the derivative of inverse trig functions, I see two different formulas given, depending upon which text you use: You can, supposedly, use the first formula, d/dx (arcsin u) = u'/(1-u^2)^1/2, or you can use the second formula, d/dx (arcsin u) = 1/(1- x^2)^1/2. (Sorry, but I don't know how to use LaTeX yet)

Don't worry about LaTeX; that's perfectly readable.

But your second formula presumably should have been d/dx (arcsin x) = 1/(1- x^2)^(1/2).

And the first formula is just the same thing, combined with the chain rule:

d/dx (arcsin u) = d/du (arcsin u) * du/dx = 1/(1- u^2)^(1/2) * u' = u'/(1-u^2)^(1/2).​

In your second work, you totally ignored the chain rule. (This is one reason they should not teach the integrated formula with the chain rule built in, because it allows you to get answers without thinking, just blindly apply a formula without paying attention to what it means.).....edited -added "not"

 

[HallsofIvy]

In learning about how to find the derivative of inverse trig functions, I see two different formulas given, depending upon which text you use: You can, supposedly, use the first formula, d/dx (arcsin u) = u'/(1-u^2)^1/2, or you can use the second formula, d/dx (arcsin u) = 1/(1- x^2)^1/2. (Sorry, but I don't know how to use LaTeX yet)

Go back and look at the texts again! The second is wrong. It should be d/dx(arcsin(x))= 1/(1- x^2)^1/2 OR d/dx(arcsin(u))= 1/(1- u^2)^1/2 (du/dx).

 

[pka]

In learning about how to find the derivative of inverse trig functions, I see two different formulas given, depending upon which text you use: You can, supposedly, use the first formula, d/dx (arcsin u) = u'/(1-u^2)^1/2, or you can use the second formula, d/dx (arcsin u) = 1/(1- x^2)^1/2.

I suspect that notation is terribly confusing.
\(\frac{d}{{dx}}\left( {\arcsin (f(x))} \right) = \frac{{f'(x)}}{{\sqrt {1 - {f^2}(x)} }}\)

 

[rayroshi]

Thanks, guys, for your responses. I really appreciate the help.

Dr. Peterson: Yes, I did indeed mean d/dx (arcsin x) = 1/(1- x^2)^(1/2), instead of d/dx (arcsin u) = 1/(1- x^2)^(1/2). Sorry. Thanks for catching that.

And you were absolutely correct regarding your comment about blindly applying a formula without understanding it. That is a very frustrating thing for me and I am totally aware of it; however, as stated, I'm trying to learn calculus without the benefit of a teacher, so am often stuck with just pushing my way through a lot of this stuff until I can then 'connect the dots'; sort of working backwards, as it were. Being a retired teacher, I'm well aware of the whole problem of 'memorizing vs. understanding,' and have preached that more times than I can count.

So the equations d/dx (arcsin u) = u'/(1-u^2)^1/2 and d/dx (arcsin x) = 1/(1- x^2)^(1/2) are both valid? But you cannot simply 'plug and chug,' as they say, with the second one?

HallsofIvy: Yes, as stated above, that was just a typo.

pka: No, the equation you displayed is not confusing to me and seems much more akin the the first equation than the second. It's the second one that doesn't seem to make sense.

 

[Dr.Peterson]

So the equations d/dx (arcsin u) = u'/(1-u^2)^1/2 and d/dx (arcsin x) = 1/(1- x^2)^(1/2) are both valid? But you cannot simply 'plug and chug,' as they say, with the second one?

Yes, they are both valid; but the first applies directly to your problem, while the second requires you to also use the chain rule. Since you need to know the chain rule for other problems, I would practice that if I were you.

I see that I somehow left out the word "not" in my parenthetical comment.

 

[skeeter]

[MATH]y = \arcsin(2x) \implies \sin{y} = 2x[/MATH]
[MATH]\dfrac{d}{dx} \bigg[\sin{y} = 2x\bigg][/MATH]
[MATH]\cos{y} \cdot \dfrac{dy}{dx} = 2[/MATH]
[MATH]\dfrac{dy}{dx} = \dfrac{2}{\cos{y}}[/MATH]
[MATH]\sin{y} = 2x \implies \cos{y} = \sqrt{1-(2x)^2}[/MATH]
[MATH]\dfrac{dy}{dx} = \dfrac{2}{\sqrt{1-(2x)^2}}[/MATH]

 

[rayroshi]

Yes, they are both valid; but the first applies directly to your problem, while the second requires you to also use the chain rule. Since you need to know the chain rule for other problems, I would practice that if I were you.

I see that I somehow left out the word "not" in my parenthetical comment.

Okay, well, thank you again for your help. As stated, I really do appreciate your time and effort.