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

In summary, to derive arcsin(1 - 2 e ^-t), use the chain rule with the derivative of arcsin being 1/√(1-x^2) and the 'u' term being 1-2e^-t. This will result in the derivative of 1-2e^-t being 1/(1-(1-2e^-t)^2) multiplied by the original derivative of arcsin. It is important to only differentiate the 'u' term and not the entire square root term when using the chain rule.
  • #1
goomer
31
0
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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  • #2
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?
 
  • #3
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]
 
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  • #4
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.
 
  • #5
Sorry, I don't know how to use the PrettyFont
1/√(1-x^2) is the derivative of arcsin, I mean
 
  • #6
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.
 
  • #7
So I am right in trying to differentiate (1 - 2 e ^-t) for the chain rule in my original problem?
 
  • #8
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:
 

1. What is the chain rule?

The chain rule is a formula that is used to find the derivative of a composite function. It is used when a function is composed of two or more functions, and you need to find the derivative of the overall function.

2. How do I know which chain rule to use?

There are different versions of the chain rule, such as the power rule, the product rule, and the quotient rule. To determine which chain rule to use, you need to identify the outer and inner functions of the composite function and then choose the appropriate rule based on the form of the function.

3. Can the chain rule be applied to arcsin(1-2e^-t)?

Yes, the chain rule can be applied to arcsin(1-2e^-t) because it is a composite function made up of two functions, arcsin(x) and 1-2e^-t. The outer function is arcsin(x) and the inner function is 1-2e^-t.

4. What is the derivative of arcsin(1-2e^-t)?

The derivative of arcsin(1-2e^-t) can be found by using the chain rule. The formula for the derivative of arcsin(x) is 1/sqrt(1-x^2), so the derivative of arcsin(1-2e^-t) is 1/sqrt(1-(1-2e^-t)^2) times the derivative of the inner function, which is -2e^-t.

5. Why is the chain rule important in deriving arcsin(1-2e^-t)?

The chain rule is important in deriving arcsin(1-2e^-t) because it allows us to find the derivative of a composite function. Without using the chain rule, it would be difficult to find the derivative of a complicated function like arcsin(1-2e^-t).

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