Deriving arcsin(1-2e^-t): What Chain Rule to Use?

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Homework Help Overview

The discussion revolves around deriving the derivative of the function arcsin(1 - 2e^-t), focusing on the application of the chain rule in calculus.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the correct application of the chain rule, questioning what variable should be differentiated and how to handle the derivative of arcsin. There is uncertainty about whether to differentiate the entire expression or just the inner function.

Discussion Status

Several participants have provided insights into the differentiation process, with some clarifying the role of the chain rule. There is ongoing exploration of the correct interpretation of the derivative and the variable involved.

Contextual Notes

Some participants express confusion about the notation and the proper application of LaTeX for mathematical expressions, which may affect the clarity of the discussion.

goomer
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1. Derive arcsin(1 - 2 e ^-t)



2. The derivative of arcsin is 1/√(1-x^2)



3. I tried using the chain rule for 1 - 2 e ^-t, but that didn't work out. What should I take the chain rule of?
 
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goomer said:
1. Derive arcsin(1 - 2 e ^-t)
That should be "Differentiate arcsin(1 - 2e^(-t))"
goomer said:
2. The derivative of arcsin is 1/√(1-x^2)



3. I tried using the chain rule for 1 - 2 e ^-t, but that didn't work out. What should I take the chain rule of?
[tex]\frac{d}{du}arcsin(u) = \frac{1}{\sqrt{1 - u^2}}[/tex]

so

[tex]\frac{d}{dx}arcsin(u) = \frac{d}{du}arcsin(u)~\frac{du}{dx}[/tex]

Can you figure out what u is here?
 
What you have done so far is almost right. The derivative of arcsin(x) is
[tex]\frac{1}{1- x^2}[/tex]
and, in [itex]arcsin(1- 2e^{-t})[/itex], [itex]x= 1- 2e^{-t}[/itex]
So the derivative of [itex]arcsin(1- 2e^{-t})[/itex]
is
[tex]\frac{1}{1- (1- 2e^{-t})^2}[/tex]
times (the chain rule) the derivative of
[tex]1- 2e^{-t}[/tex]
 
Last edited by a moderator:
I'm still not quite sure...is it the chain rule of the entire √(1-u^2) portion?
I'm certain the derivative of arcsin is \frac{1}{\sqrt{1 - x^2}}[/tex] however.
 
Sorry, I don't know how to use the PrettyFont
1/√(1-x^2) is the derivative of arcsin, I mean
 
goomer said:
I'm still not quite sure...is it the chain rule of the entire √(1-u^2) portion?
I'm certain the derivative of arcsin is \frac{1}{\sqrt{1 - x^2}}[/tex] however.

When applying the chain rule, you only differentiate the 'u' term by itself, not the whole square root term. See Mark44's post.
 
So I am right in trying to differentiate (1 - 2 e ^-t) for the chain rule in my original problem?
 
goomer said:
I'm still not quite sure...is it the chain rule of the entire √(1-u^2) portion?
I'm certain the derivative of arcsin is \frac{1}{\sqrt{1 - x^2}}[/tex] however.

You'll need to put [ tex ] in front of your equation to (without spaces). So you should have typed [ tex ]\frac{1}{\sqrt{1 - x^2}}[ /tex ].
That's how you use the PrettyFont here (called LaTeX). Alternatively, you can also use the x2 and x2 buttons above to create su(b/p)scripts.

Sorry, just wanted to mention that :biggrin:
 

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