Solving Derivative y=√x+√x Homework

[tinala]

Homework Statement

derivative y=√x+√x

Homework Equations

The Attempt at a Solution

 

[Kevin_Axion]

\sqrt{x} = x^{\frac{1}{2}

 

[tinala]

\sqrt{x} = x^{\frac{1}{2}

yes but y=sqrt x+sqrtx

 

[Char. Limit]

I assume you mean

y=\sqrt{x+\sqrt{x}}

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:

y=\sqrt{x}+\sqrt{x}

 

[tinala]

I assume you mean

y=\sqrt{x+\sqrt{x}}

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:

y=\sqrt{x}+\sqrt{x}

yes that is what I mean but I'm stuck so please would you help me ?

 

[Char. Limit]

Well, do you know the chain rule?

 

[tinala]

no we haven't studied that yet

 

[Char. Limit]

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...

\frac{df}{dx} = \frac{df}{dg} \frac{dg}{dx}

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

 

[tinala]

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...

\frac{df}{dx} = \frac{df}{dg} \frac{dg}{dx}

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

thank you very much