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Find the derivative of $y=\sqrt{1+\sqrt{x}}$

I know it comes out to $$y' = \frac {1} {4\sqrt{x}\sqrt{1+\sqrt{x}}}$$

But have no ideas on how to get there or where to start.

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5 Answers

up vote 2 down vote accepted

Do you know the chain rule? Usually when you have to take a derivative of something of the form $f(g(x))$, that's where you start. The chain rule says that the derivative is $$f'(g(x))\cdot g'(x).$$

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Hint $$(f\circ g)'=(f'\circ g)\times g'$$ and $$(\sqrt{x})'=\frac{1}{2\sqrt{x}}$$

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Alternatively, you can rearrange to get $$x=(y^2-1)^2$$Then differentiate both sides and resolve $y'$. Of course you will still need to apply chain rule.

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$y=\sqrt{1+\sqrt x}$

$y=\sqrt{u}$ [Let $u=1+\sqrt x$]

$\displaystyle y'=\frac 1 {2\sqrt u}\frac 1 {2\sqrt x}$ [$\displaystyle\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$]

$\displaystyle y'=\frac 1 {4\sqrt x\sqrt{1+\sqrt x}}$

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As MJD says, you use the chain rule as follows.\begin{align*}&y=\sqrt{1+\sqrt x}=(1+\sqrt x)^{1/2}\\\Rightarrow&y'=\frac{1}{2}(1+\sqrt x)^{-1/2}\cdot \frac {1}{2\sqrt x}\\&y' =\frac{1}{4\sqrt x\sqrt{1+\sqrt x}},\end{align*}which is what you want.

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