Derivatives of Square Root Functions: Understanding the Chain Rule

[unf0r5ak3n]

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]

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

 

[hotvette]

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 = x^1/2.

 

[unf0r5ak3n]

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

sorry I meant equivalent to

 

[mathman]

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]

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

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

\frac{dy}{dx}= \frac{dy}{du}\frac{du}{dx}

 

[unf0r5ak3n]

if u = 1-sin(x) then du/dx = (1/2)u

 

[brewAP2010]

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)

 

[HallsofIvy]

if u = 1-sin(x) then du/dx = (1/2)u

?? I was under the impression that the derivative of 1- sin(x) was -cos(x).