Hello everyone. This lesson we will tell the derivative of $\sqrt{1+\sqrt{1+\sqrt{x}}}$.

## What is the derivative of $\sqrt{1+\sqrt{1+\sqrt{x}}}$?

**Compute the derivative of $\sqrt{1+\sqrt{1+\sqrt{x}}}$.**

Here we have a more complicated chain of compositions, so we use the chain rule twice. At the outermost “layer” we have the function $g(x)=1+\sqrt{1+\sqrt{x}}$ plugged into $f(x)=\sqrt{x}$, so applying the chain rule once gives

$\frac{d}{dx}\sqrt{1+\sqrt{1+\sqrt{x}}}=\frac{1}{2}(1+\sqrt{1+\sqrt{x}})^{\frac{-1}{2}}.\frac{d}{dx}(1+\sqrt{1+\sqrt{x}})$

**Not we need the derivative**of $\sqrt{1+\sqrt{x}}$. Using the chain rule again:

$\frac{d}{dx}\sqrt{1+\sqrt{x}}=\frac{1}{2}(1+\sqrt{x})^{\frac{-1}{2}}.\frac{1}{2}x^{\frac{-1}{2}}$

**So the original derivative is**

$\frac{d}{dx}\sqrt{1+\sqrt{1+\sqrt{x}}}=\frac{1}{2}(1+\sqrt{1+\sqrt{x}})^{\frac{-1}{2}}.\frac{1}{2}(1+\sqrt{x})^{\frac{-1}{2}}.\frac{1}{2}x^{\frac{-1}{2}}$

=$\frac{1}{8\sqrt{x}.\sqrt{1+\sqrt{x}}.\sqrt{1+\sqrt{1+\sqrt{x}}}}$

Using the chain rule, the power rule, and the product rule, it is possible to avıid using the quotient rule entirely.

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