Derivative y=√x+√x

  • Thread starter tinala
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  • #1
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Homework Statement



derivative y=√x+√x

Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
901
2


[tex]\sqrt{x} = x^{\frac{1}{2}[/tex]
 
  • #3
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[tex]\sqrt{x} = x^{\frac{1}{2}[/tex]

yes but y=sqrt x+sqrtx
 
  • #4
Char. Limit
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I assume you mean

[tex]y=\sqrt{x+\sqrt{x}}[/tex]

If that's the case, use the chain rule. But be sure to mark clearly with parentheses. What you wrote could just as easily be:

[tex]y=\sqrt{x}+\sqrt{x}[/tex]
 
  • #5
8
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I assume you mean

[tex]y=\sqrt{x+\sqrt{x}}[/tex]

If that's the case, use the chain rule. But be sure to mark clearly with parentheses. What you wrote could just as easily be:

[tex]y=\sqrt{x}+\sqrt{x}[/tex]

yes that is what I mean but I'm stuck so please would you help me ?
 
  • #6
Char. Limit
Gold Member
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Well, do you know the chain rule?
 
  • #7
8
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no we havent studied that yet
 
  • #8
Char. Limit
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Well, that's not good, because you need the chain rule to solve this. Basically, it states that if we have two functions, f(x) and g(x), that the derivative of f(g(x)) is f'(g(x)*g'(x). Or, in Leibniz notation...

[tex]\frac{df}{dx} = \frac{df}{dg} \frac{dg}{dx}[/tex]

Now, by setting [itex]f(g(x)) = \sqrt(g(x))[/itex] and [itex]g(x)=x+\sqrt(x)[/itex], you can use the chain rule to get the derivative.
 
  • #9
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Well, that's not good, because you need the chain rule to solve this. Basically, it states that if we have two functions, f(x) and g(x), that the derivative of f(g(x)) is f'(g(x)*g'(x). Or, in Leibniz notation...

[tex]\frac{df}{dx} = \frac{df}{dg} \frac{dg}{dx}[/tex]

Now, by setting [itex]f(g(x)) = \sqrt(g(x))[/itex] and [itex]g(x)=x+\sqrt(x)[/itex], you can use the chain rule to get the derivative.

thank you very much
 

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