- #1

tinala

- 8

- 0

## Homework Statement

derivative y=√x+√x

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- Thread starter tinala
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- #1

tinala

- 8

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derivative y=√x+√x

- #2

Kevin_Axion

- 912

- 2

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

- #3

tinala

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- #4

Char. Limit

Gold Member

- 1,216

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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

tinala

- 8

- 0

[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

tinala

- 8

- 0

no we havent studied that yet

- #8

Char. Limit

Gold Member

- 1,216

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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

tinala

- 8

- 0

[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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