Chain rule for square roots (1 Viewer)

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I'm kind of confused about how to approach a function with the chain rule.

For example in the equation ƒ(x) = sqrt(1-sin(x)) I know i simplify it to ƒ(x) = 1-sin(x)^(1/2) but I'm lost from there.
 

mathman

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What are you trying to accomplish? The "simplification" you give doesn't make any sense.
 

hotvette

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I'm kind of confused about how to approach a function with the chain rule.

For example in the equation ƒ(x) = sqrt(1-sin(x)) I know i simplify it to ƒ(x) = 1-sin(x)^(1/2) but I'm lost from there.
From the title of your post I'm guessing you want to find f'(x). Your simplification isn't quite correct. Don't you mean ƒ(x) = (1-sin(x))^(1/2)? Fractional powers are treated the same as integer powers when it comes to differentiation. I presume you know how to differentiate y = x3. Use the exact same methodology to differentiate y = x1/2.
 

mathman

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sorry I meant equivalent to
I don't know what "equivalent to" means either in this context. However sqrt(1-sin(x)) is just different from 1-sin(x)^(1/2). There is no way to equate these two expressions.
 

HallsofIvy

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The point being that [itex]1- sin(x)^{1/2}\ne (1- sin(x))^{1/2}[/itex].

Write [itex](1- sin(x))^{1/2}[/itex] as [itex]y= u^{1/2}[/itex] with [itex]u= 1- sin(x)[/itex]. Can you find dy/du and du/dx?

[tex]\frac{dy}{dx}= \frac{dy}{du}\frac{du}{dx}[/tex]
 
if u = 1-sin(x) then du/dx = (1/2)u
 
If f(x)= u^n then f’(x)= (n)u^(n-1)u’
so
If f(x)= (1-sinx)^1/2 then f’(x)= (1/2)(1-sinx)^(-1/2)(-cosx)
 

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