Master the Chain Rule: Derivative of f(x) = \sqrt{5-x^{2}} and Its Composition

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Homework Statement



A: Write f(x) = [tex]\sqrt{5-x^{2}}[/tex] as a composite of two functions.

B: Use the Chain Rule to find the derivative of f(x) = [tex]\sqrt{5-x^{2}}[/tex]

Homework Equations



Chain Rule:

y`= [tex]\frac{dy}{du}[/tex] [tex]\frac{du}{dx}[/tex]

The Attempt at a Solution



A:

y = [tex]\sqrt{u}[/tex]
u = 5 - x[tex]^{2}[/tex]


B:

This is where I get confused. I don't understand what's meant by "d" and what's meant by "y", "u", and "x".

I know the two du's cancel out in the Chain Rule so you're left with:

y`= (dy)(dx)

Does this mean the derivative of y times the derivative of x? And if so, how do you know what y and x are?
 
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Incog said:
y = [tex]\sqrt{u}[/tex]
u = 5 - x[tex]^{2}[/tex]
Good!
This is where I get confused. I don't understand what's meant by "d" and what's meant by "y", "u", and "x".
dy/dx is just notation for: take the derivative of the function y with respect to the variable x.

I know the two du's cancel out in the Chain Rule so you're left with:

y`= (dy)(dx)
Not exactly. y' = dy/dx.

The point is that you want to find dy/dx and that dy/dx = (dy/du)*(du/dx). And it's easy to calculate the derivatives dy/du and du/dx.
 
Doc Al said:
dy/dx is just notation for: take the derivative of the function y with respect to the variable x.

Thanks. That cleared the confusion. :cool: