Derivatives of Square Root Functions: Understanding the Chain Rule

In summary, the conversation discusses the confusion of approaching a function with the chain rule, specifically the equation ƒ(x) = sqrt(1-sin(x)). The individual is attempting to find f'(x) and is unsure of how to simplify the equation. The expert clarifies the proper way to simplify and differentiate using fractional powers and provides an example. The conversation also touches on the concept of equivalency between expressions and differentiating using the power rule.
  • #1
unf0r5ak3n
36
0
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.
 
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  • #2
What are you trying to accomplish? The "simplification" you give doesn't make any sense.
 
  • #3
unf0r5ak3n said:
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.
 
  • #4
mathman said:
What are you trying to accomplish? The "simplification" you give doesn't make any sense.

sorry I meant equivalent to
 
  • #5
unf0r5ak3n said:
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.
 
  • #6
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]
 
  • #7
if u = 1-sin(x) then du/dx = (1/2)u
 
  • #8
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)
 
  • #9
unf0r5ak3n said:
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).
 

What is the chain rule for square roots?

The chain rule for square roots is a rule in calculus that allows us to find the derivative of a function that involves a square root. It is used when the function is composed of two or more functions, with one of them being a square root function.

Why is the chain rule important for square roots?

The chain rule is important for square roots because it allows us to find the rate of change of a function that involves a square root. This is useful in real-world applications, such as in physics and engineering, where many functions involve square roots.

What is the formula for the chain rule for square roots?

The formula for the chain rule for square roots is:

f'(x) = (1/2) * (1/sqrt(g(x))) * g'(x)

Where f(x) is the outer function and g(x) is the inner function.

How is the chain rule for square roots applied in practice?

The chain rule for square roots is applied by first identifying the inner and outer functions in a given function. Then, we use the formula to find the derivative of the outer function, and finally, multiply it by the derivative of the inner function. This gives us the derivative of the original function.

Can the chain rule be applied to other types of roots?

Yes, the chain rule can be applied to other types of roots, such as cube roots, fourth roots, and so on. The only difference is that the power in the formula will change accordingly. For example, for cube roots, the power would be (1/3) instead of (1/2) as in the case of square roots.

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