Question

Don't understand how to simplified this:
frac{ \sqrt{x^2+1}+x }{ \sqrt{x^2+1} (the bracket is suppose between the second one)
Answer:
\[= \frac{ \sqrt{x+\sqrt{x^2+1}} }{ 2\sqrt{x^2+1} }\]

Answer 1

it mean this:
not sure to the simplify this into that answer:
\[y'= \frac{ 1 }{ 2\sqrt{x+\sqrt{x^2+1}} } \left[ \frac{ \sqrt{x^2+1}+x }{ \sqrt{x^2+1} } \right]\]
Sorry~ (it was hard for me type it up)
\[= \frac{ \sqrt{x+\sqrt{x^2+1}} }{ 2\sqrt{x^2+1} }\]

Answer 2

You have
\[y'= \frac{ 1 }{ 2\color{blue}{\sqrt{x+\sqrt{x^2+1}}} } \left[ \frac{ \sqrt{x^2+1}+x }{ \sqrt{x^2+1} } \right]\]

Multiply top and bottom by \(\color{blue}{\sqrt{x+\sqrt{x^2+1}}}\), what do you get ?

Answer 3

I'm not sure how to do that...

Answer 4

do you know how to rationalize the denominator of \(\large \dfrac{a}{\sqrt{b}}\) ?

Answer 5

yoy multiply top and bottom to get rid of the square root

Answer 6

but I'm don't know what to do with double square roots

Answer 7

we use the same idea here

Answer 8

\[\begin{align}y'&= \frac{ 1 }{ 2\color{blue}{\sqrt{x+\sqrt{x^2+1}}} } \left[ \frac{ \sqrt{x^2+1}+x }{ \sqrt{x^2+1} } \right]\\~\\
&= \frac{ 1 }{ 2\color{blue}{\sqrt{x+\sqrt{x^2+1}}} } \left[ \frac{ \sqrt{x^2+1}+x }{\sqrt{x^2+1} } \right] \times\dfrac{\color{blue}{\sqrt{x+\sqrt{x^2+1}}} }{\color{blue}{\sqrt{x+\sqrt{x^2+1}}} } \\~\\
&= \frac{ 1 }{ 2 } \left[ \frac{ \sqrt{x^2+1}+x }{\sqrt{x^2+1} } \right] \times\dfrac{\color{blue}{\sqrt{x+\sqrt{x^2+1}}} }{\left(\color{blue}{\sqrt{x+\sqrt{x^2+1}}} \right)^2} \\~\\
\end{align}\]

Answer 9

square and sqrt cancel each other :

\[\require{cancel}\begin{align}y'&= \frac{ 1 }{ 2\color{blue}{\sqrt{x+\sqrt{x^2+1}}} } \left[ \frac{ \sqrt{x^2+1}+x }{ \sqrt{x^2+1} } \right]\\~\\
&= \frac{ 1 }{ 2\color{blue}{\sqrt{x+\sqrt{x^2+1}}} } \left[ \frac{ \sqrt{x^2+1}+x }{\sqrt{x^2+1} } \right] \times\dfrac{\color{blue}{\sqrt{x+\sqrt{x^2+1}}} }{\color{blue}{\sqrt{x+\sqrt{x^2+1}}} } \\~\\
&= \frac{ 1 }{ 2 } \left[ \frac{ \sqrt{x^2+1}+x }{\sqrt{x^2+1} } \right] \times\dfrac{\color{blue}{\sqrt{x+\sqrt{x^2+1}}} }{\left(\color{blue}{\sqrt{x+\sqrt{x^2+1}}} \right)^2} \\~\\
&= \frac{ 1 }{ 2 } \left[ \frac{ \sqrt{x^2+1}+x }{\sqrt{x^2+1} } \right] \times\dfrac{\color{blue}{\sqrt{x+\sqrt{x^2+1}}} }{\color{blue}{x+\sqrt{x^2+1}}} \\~\\
&= \frac{ 1 }{ 2 } \left[ \frac{ \cancel{\sqrt{x^2+1}+x }}{\sqrt{x^2+1} } \right] \times\dfrac{\color{blue}{\sqrt{x+\sqrt{x^2+1}}} }{\color{blue}{\cancel{x+\sqrt{x^2+1}}}} \\~\\
\end{align}\]

Answer 10

Wow! Thank you! Did not that you can reduce that. I tried changing the square roots with exponents. silly me..

Answer 11

where did you get the square on the bottom denominator?

Answer 12

nevermind , i got it

Answer 13

i see what you did that! Thanks!

Answer 14

\[\large \color{blue}{\sqrt{\heartsuit}} \times \color{blue}{\sqrt{\heartsuit}} = (\color{blue}{\sqrt{\heartsuit}} )^2\]

Answer 15

Thank you! :D