Question

F(x)=√(x+√(x²+1))
calculate Df
Prove that
(∀x∈IR):4(1+x^2 ) f^'' (x)+4xf^' (x)-f(x)=0

Answer 1

\[f(x)=\sqrt{x+\sqrt{x²+1}}\]

Answer 2

\[F'(x)=\frac{ d }{ dx }\left\{ x+\left( x^2+1 \right)^{\frac{ 1 }{ 2 }} \right\}^{\frac{ 1 }{ 2 }}\]
\[=\frac{ 1 }{2 }\left\{ x+(x^2+1)^{\frac{ 1 }{ 2 }} \right\}^{\frac{ -1 }{ 2 }}\left[ 1+\frac{ 1 }{2 }(x^2+1)^{\frac{ -1 }{ 2 }}*2x \right] \]
simplify it.

Answer 3

What is d ?

Answer 4

\(\dfrac d{dx}\) means differentiate with respect to x.

Answer 5

\[f'(x)=\frac{ 1 }{2 }\frac{ 1 }{ \sqrt{x+\sqrt{x^2+1}} }\left[ 1+\frac{ x }{ \sqrt{x^2+1} } \right]\]
\[=\frac{ 1 }{2 }\frac{ 1 }{\sqrt{x+\sqrt{x^2+1}} }\frac{ x+\sqrt{x^2+1} }{\sqrt{x^2+1} }\]
\[=\frac{ 1 }{2 }\frac{ \sqrt{x+\sqrt{x^2+1}} }{\sqrt{x^2+1} }=\frac{ f(x) }{2\sqrt{x^2+1} } ....(1) \]
\[2\sqrt{x^2+1}f'(x)=f(x)\]
again differentiate
\[2\sqrt{x^2+1}f \prime \prime(x)+2f \prime(x)\frac{ 2x }{2\sqrt{x^2+1} }=f \prime(x)\]
\[2\sqrt{x^2+1}f \prime \prime(x)+\frac{ 2x }{\sqrt{x^2+1} }f \prime(x)=\frac{ f(x) }{2\sqrt{x^2+1} }~from~(1)\]
\[multiply~by~2\sqrt{x^2+1}\]
\[4(x^2+1)f \prime \prime(x)+4xf \prime(x)=f(x)\]
\[or~4(1+x^2)f \prime \prime(x)+4xf \prime(x)-f(x)=0\]