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 .

  • $\begingroup$ In this case, it is better to wait until you speak of the chain rule. It won't take long. $\endgroup$ – Alex M. Sep 24 '16 at 19:03
  • $\begingroup$ Have you seen implicit differentiation? if $y=\sqrt {a+\sqrt x}$ then $y^2= a+\sqrt x$ and it is easy to differentiate both sides. $\endgroup$ – lulu Sep 24 '16 at 19:04
  • $\begingroup$ Mild terminology comment: it's differentiate, not derive. $\endgroup$ – Sean Roberson Sep 24 '16 at 20:07
  • $\begingroup$ @lulu: The OP explicitly said that (s)he hasn't studied the chain rule yet, so I guess that implicit differentiation is a bit streched already. $\endgroup$ – Alex M. Sep 24 '16 at 21:19
  • $\begingroup$ It's a peace of cake: apply your square root rule with $u=a+\sqrt a$. $\endgroup$ – Michael Hoppe Sep 24 '16 at 22:59

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 $y$:

$$y'y+yy'=2yy'=\frac1{2\sqrt u}$$

$$y'=\frac1{4y\sqrt u}$$

  • $\begingroup$ Another way to write the chain rule that may be clearer: $$ \frac{\mathrm{d}}{\mathrm{d}u} f(g(u)) = g'(u)f'(g(u))$$ $\endgroup$ – Zubin Mukerjee Sep 24 '16 at 20:07
  • $\begingroup$ @ZubinMukerjee Ah, thanks for the new fonts. $\endgroup$ – Simply Beautiful Art Sep 24 '16 at 20:28
  • $\begingroup$ No problem, I think that is just a stylistic preference. But I do think adding the argument of the functions ($g(u)$ as opposed to just $g$) does increase clarity :) Good answer, regardless +1 $\endgroup$ – Zubin Mukerjee Sep 24 '16 at 20:31
  • $\begingroup$ @ZubinMukerjee Also true. Probably should've used it here considering the OP. $\endgroup$ – Simply Beautiful Art Sep 24 '16 at 20:32
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    $\begingroup$ @AlexM. Well, sure. On the other hand, the chain rule for $y^2$ is a complete triviality and that's all that's required here. $\endgroup$ – lulu Sep 24 '16 at 21:30

$$y=\root \of{ a + \root \of u}$$ $$y'=\frac{1}{2}(a+\sqrt{u})^{\frac{1}{2}-1}(\sqrt{}u)'$$ $$(\sqrt{u})'=\frac{1}{2}u^{\frac{1}{2}-1}$$

  • $\begingroup$ The OP explicitly said that (s)he hasn't studied the chain rule yet. $\endgroup$ – Alex M. Sep 24 '16 at 21:18

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