# Derivative y=√x+√x

tinala

## Homework Statement

derivative y=√x+√x

## Answers and Replies

Kevin_Axion

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

tinala

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

yes but y=sqrt x+sqrtx

Gold Member

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 ?

Gold Member

Well, do you know the chain rule?

tinala

no we havent studied that yet

Gold Member

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