The derivative of g(x)=R(sqrt(u(x))?

  • Thread starter Cuisine123
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  • #1
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Homework Statement


Find the derivative of g(x)=R(sqrt(u(x)). Both R and u are differentiable functions.

Homework Equations


N/A


The Attempt at a Solution


I have been told that the solution is R' (sqrt(u(x)) * (1/2)u^(-1/2) * u' but I have no idea why. Please explain how to get to this solution.
Thanks.
 

Answers and Replies

  • #2
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are you familiar with the chain rule?
 
  • #3
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are you familiar with the chain rule?

Yes I am. Please tell me how to proceed.
 
  • #4
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well if you know the chain rule, that's where you start. how would you write the chain rule for this problem?
 
  • #5
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well if you know the chain rule, that's where you start. how would you write the chain rule for this problem?

I have tried using the chain rule but I don't know how to apply it to this question though.
 
  • #6
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well the basic idea is that you have a function that is composed of many different functions. For example [tex]\sin(e^{x^{2}}), [/tex]

by calling [tex] u(x)=x^{2} , v(u)=e^{u}, w(v)=\sin(v) [/tex]

we can rewrite this complicated mess as [tex] w( v ( u (x) ) ) [/tex]

and from the chain rule , we get

[tex] w'(x)=w'( v ( u(x) ) ) *v'( u(x) ) *(u'(x)) [/tex]

in your problem ,[tex] W(v) = R(v) , v(u)=\sqrt{u} , u(x)=u(x). [/tex]

see if you can use that
 
  • #7
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well the basic idea is that you have a function that is composed of many different functions. For example [tex]\sin(e^{x^{2}}), [/tex]

by calling [tex] u(x)=x^{2} , v(u)=e^{u}, w(v)=\sin(v) [/tex]

we can rewrite this complicated mess as [tex] w( v ( u (x) ) ) [/tex]

and from the chain rule , we get

[tex] w'(x)=w'( v ( u(x) ) ) *v'( u(x) ) *(u'(x)) [/tex]

in your problem ,[tex] W(v) = R(v) , v(u)=\sqrt{u} , u(x)=u(x). [/tex]

see if you can use that

I am getting even more confused. :cry:
 
  • #8
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Homework Statement


Find the derivative of g(x)=R(sqrt(u(x)). Both R and u are differentiable functions.

You will first use the product rule and then use the power and chain rule together to find the derivative. To start, using the product rule,
[tex]g'(x) = R'\sqrt{u(x)} + R [\sqrt{u(x)}]'[/tex]
Now, can you find [itex][\sqrt{u(x)}]'[/itex] using the chain rule?

By the way, your first * in your solution should be a +.
 
  • #9
Fredrik
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I think you just need to take a deep breath and start reminding yourself of the things you already know. I mean "obvious" things like: What is a function? What does the notation v'(u(x)) mean. v and v' are both functions. v' is the derivative of v. v'(u(x)) is a number, it's the value of v' at u(x). When you have all those things sorted out, the rest is very easy.

Can you see what function v' is?
 
  • #10
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You will first use the product rule and then use the power and chain rule together to find the derivative. To start, using the product rule,
[tex]g'(x) = R'\sqrt{u(x)} + R [\sqrt{u(x)}]'[/tex]
Now, can you find [itex][\sqrt{u(x)}]'[/itex] using the chain rule?

By the way, your first * in your solution should be a +.

[itex][\sqrt{u(x)}]'[/itex] using the chain rule, I get 1/2([tex]\sqrt{}u(x))[/tex]
 
  • #11
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[itex][\sqrt{u(x)}]'[/itex] using the chain rule, I get 1/2([tex]\sqrt{}u(x))[/tex]

Well you didn't use the power rule correctly, and you didn't use the chain rule. For example,
[tex][(u(x))^3]' = 3(u(x)^2)\cdot u'(x)[/tex]
You have to use the power rule to take the derivative of the "outside", and then multiply by the derivative of the "inside", which is u'(x). This is how the chain rule applies. If this doesn't make sense, then you need to review the section on the chain rule in your text.

Can you apply this knowledge to finish out the problem now?
 
  • #12
Fredrik
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I really don't like the notation [tex][\sqrt{u(x)}]'[/tex]. When you see something with a prime on it, like G', it's supposed to refer to a function called G' which is the derivative of another function, called G. (And G'(y) is supposed to be a number, the value of G' at the point y). But [tex]\big[\sqrt{u(x)}\big][/tex] is a number, not a function. The function you have in mind is written as [tex]x\mapsto\sqrt{u(x)}[/tex]. There is a standard way to write "the derivative of [tex]x\mapsto\sqrt{u(x)}[/tex], evaluated at x", and it looks like this:

[tex]\frac{d}{dx}\sqrt{u(x)}[/tex]

But this notation gets a bit awkward when you have to use the chain rule, especially when you don't have a lot of experience with it. So I recommend that you first express the function [tex]x\mapsto\sqrt{u(x)}[/tex] in a different way, by first defining a function g by [tex]g(x)=\sqrt{x}[/tex] for all x. (Note that it makes no difference whatsoever what I call the independent variable in this definition, because I'm defining a function. If I define h by [tex]h(t)=\sqrt{t}[/tex] for all t, then h=g). Now you have

[tex]\frac{d}{dx}\sqrt{u(x)}=\frac{d}{dx}g(u(x))=\frac{d}{dx}(g\circ u)(x)=(g\circ u)'(x)[/tex]

Can you use the chain rule now? Recall that the chain rule says

[tex](g\circ u)'(x)=g'(u(x))u'(x)[/tex].

To evaluate the right-hand side, you need to be able to answer all of the following:

1. What is the derivative of u? (The answer is a function u', not a number).
2. What is the derivative of g? (The answer is a function g', not a number).
3. What is the value of u' at the point x? (In other words, what number do you get when you insert the number x into the function u').
4. What is the value of u at the point x?
5. What is the value of g' at the point u(x)? (In other words, what number do you get when you insert the number u(x) into the function g').

This is why I recommended that you really think about the basics, like which notations represent functions, which ones represent numbers, and so on. It gets really easy when you have those details figured out.
 

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