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
$frac{d}{dx}sqrt{1+sqrt{1+sqrt{x}}}=frac{1}{2}(1+sqrt{1+sqrt{x}})^{frac{-1}{2}}.frac{1}{2}(1+sqrt{x})^{frac{-1}{2}}.frac{1}{2}x^{frac{-1}{2}}$
=$frac{1}{8sqrt{x}.sqrt{1+sqrt{x}}.sqrt{1+sqrt{1+sqrt{x}}}}$
Using the chain rule, the power rule, and the product rule, it is possible to avıid using the quotient rule entirely.
