We've just started learning deriviation of a single variable function, I really like the subject and I feel comfortable deriving most kind of polynomial, squared root func and other types (except for $ln$ , $log$ and $e^x$ kind of functions ).
Anyways, I was wondering how can we derive a square root function with another square root anside of it .
For example let's say that $u$ is a func, and $a$ is a real number, how can we derive this function . :
$$\root \of{ a + \root \of u}$$
Or this, given that $v$ is a function :
$$\root \of{ v + \root \of u}$$
I know that in order to derive a square root function we apply this :
$$(\root \of u) ' = \frac{u '}{2\root \of u}$$
But I really can't find a way on how to do the first two function derivatives, I've heard about the chain rule, but we didn't use it yet .
Thank's for your time .
Best Answer
We apply chain rule.
$$\frac{\mathrm{d}}{\mathrm{d}u} f(g(u)) = g'(u)f'(g(u))$$
For your case, $f(u)=\sqrt{a+u}$ and $g(u)=\sqrt u$. Plugging these in, we get:
$$\frac{\mathrm{d}}{\mathrm{d}u}\sqrt{a+\sqrt u}=\frac1{4\sqrt u\sqrt{a+\sqrt u}}=\frac1{4\sqrt{au+u\sqrt u}}$$
Or, you could manipulate as follows:
$$y^2=y\times y=a+\sqrt u$$
$$\frac{\mathrm{d}}{\mathrm{d}u}y\times y=\frac1{2\sqrt u}$$
Apply product rule and solve for $y$:
$$y'y+yy'=2yy'=\frac1{2\sqrt u}$$
$$y'=\frac1{4y\sqrt u}$$