Question

y=sqrt(x+sqrt(x²+1)), differentiate using chain rule

Answer 1

\[y=\sqrt{x \sqrt{x²+1}}\]

Answer 2

\[y' = \frac{ 1 }{ 2\sqrt{x+\sqrt{x^2+1}} }*1*\frac{ 1 }{ 2\sqrt{x^2+1} }*2x\]

Answer 3

so I get three functions \[f(x)=\sqrt{x}, g(x)=\sqrt{x}, h(x)=x²+1\]

Answer 4

what's the 1 in the middle?

Answer 5

I peel it like an onion: work from the outside to the inside, everytime multiplying. Works well, and you don't need to worry about the names of all intermediate functions.

So: \(y'=\dfrac{1}{2\sqrt{x+\sqrt{x^2+1}}}\cdot \left(1+\dfrac{1}{2\sqrt{x^2+1}}\right)\cdot 2x\).

Answer 6

Now you may have to simplify the whole thing, but that is only algebra...

Answer 7

\[=2x+\frac{ 2x }{ 2\sqrt{x²+1} }*\frac{ 1 }{ 2\sqrt{x+\sqrt{x²+1}} }\]

Answer 8

\[=\frac{ 2x+\frac{ 2x }{ 2\sqrt{x²+1} } }{ 2\sqrt{x+\sqrt{x²+1}} }\]

Answer 9

what then?...

Answer 10

\[=2x+\frac{ 2\sqrt{x \sqrt{x²+1}} }{ 2\sqrt{x²+1} }*2x\] ?

Answer 11

...which seems to be close to what wolfram alpha gives, but there is still some extra in my equation

Answer 12

First note the the "bottom-left 2" and the "rightmost 2" in my formula cancel.
Also, that last factor x can be used as numerator in the first fraction:

\(\dfrac{x}{\sqrt{x+\sqrt{x^2+1}}}\cdot \left( 1+\dfrac{1}{2\sqrt{x^2+1}}\right)=\\ \dfrac{x}{\sqrt{x+\sqrt{x^2+1}}}\cdot \left( \dfrac{2\sqrt{x^2+1}}{2\sqrt{x^2+1}}+\dfrac{1}{2\sqrt{x^2+1}}\right)=\)
\(\dfrac{x}{\sqrt{x+\sqrt{x^2+1}}}\cdot \dfrac{1+2\sqrt{x^2+1}}{2\sqrt{x^2+1}}=\)
\(\dfrac{x(1+2\sqrt{x^2+1})}{2\sqrt{x^2+1}\cdot \sqrt{x+\sqrt{x^2+1}}}\)

I think this should do it :)

Answer 13

why is the first derivative \[\frac{ x }{ \sqrt{x+\sqrt{x²+1}} }\] and not \[\frac{ 1 }{ \sqrt{x+\sqrt{x²+1}} }\] ?
And why is the d(x²+1)/dx=2x now missing?

Answer 14

That is not the 1st derivative anymore, and 2x is not missing.
I have simplified the original y' (see my differentiation, and the next post in which the simplification is done).