Simplifying this derivative....

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SUMMARY

The discussion centers on evaluating the derivative of the function f(w) = cos(sin^(-1)(2w)). The correct derivative, as provided by the textbook, is F'(w) = (-4w)/sqrt(1-4w^(2)). The user initially applied the chain rule correctly but was confused about the simplification involving the inverse sine function. The key takeaway is that sin(sin^(-1)(x)) simplifies directly to x, which explains the disappearance of the sin and sin^(-1) terms in the final derivative.

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  • Familiarity with inverse trigonometric functions
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Jess Karakov
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Homework Statement


Evaluate the derivative of the following function:
f(w)= cos(sin^(-1)2w)

Homework Equations


Chain Rule
eq0008M.gif


The Attempt at a Solution


I did just as the chain rule says where
F'(w)= -[2sin(sin^(-1)2w)]/[sqrt(1-4w^(2))

but the book gave the answer as F'(w)=(-4w)/sqrt(1-4w^(2))

and I'm wondering if there is some kind of simplification I'm missing. Why would sin and sin^(-1) disappear?
 
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Jess Karakov said:

Homework Statement


Evaluate the derivative of the following function:
f(w)= cos(sin^(-1)2w)

Homework Equations


Chain Rule
View attachment 206165

The Attempt at a Solution


I did just as the chain rule says where
F'(w)= -[2sin(sin^(-1)2w)]/[sqrt(1-4w^(2))

but the book gave the answer as F'(w)=(-4w)/sqrt(1-4w^(2))

and I'm wondering if there is some kind of simplification I'm missing. Why would sin and sin^(-1) disappear?

Because they are inverse functions. ##\sin(\sin^{-1}(x))=x##.
 
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Dick said:
Because they are inverse functions. ##\sin(\sin^{-1}(x))=x##.
Ah okay. I wasn't aware of this. Thank you.
 

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