Question

Derivatives question:

Answer 1

\[y = (x+ \sqrt{x^2 +1 })^4\]

Answer 2

Prove that :
\[(x^2 + 1)y ^{''} + xy ^{'} = 16y\]

Answer 3

here we have to compute the first and second derivative of y(x)

Answer 4

It got complicated much !

Answer 5

no, it is very simple, please note that:
\[\Large y' = \frac{{4y}}{{\sqrt {{x^2} + 1} }}\]

Answer 6

Oo, But how did you get rid of 4 power ?

Answer 7

Hey ?

Answer 8

here are your steps:
\[\Large \begin{gathered}
y' = 4{\left\{ {x + {{\left( {{x^2} + 1} \right)}^{1/2}}} \right\}^3}\left\{ {1 + \frac{x}{{{{\left( {{x^2} + 1} \right)}^{1/2}}}}} \right\} = \hfill \\
\hfill \\
= 4{\left\{ {x + {{\left( {{x^2} + 1} \right)}^{1/2}}} \right\}^3}\frac{{x + {{\left( {{x^2} + 1} \right)}^{1/2}}}}{{{{\left( {{x^2} + 1} \right)}^{1/2}}}} = \hfill \\
\hfill \\
= 4\frac{{{{\left\{ {x + {{\left( {{x^2} + 1} \right)}^{1/2}}} \right\}}^4}}}{{{{\left( {{x^2} + 1} \right)}^{1/2}}}} = \frac{{4y}}{{\sqrt {{x^2} + 1} }} \hfill \\
\end{gathered} \]

Answer 9

Wow ! I didn't think about the 2nd step. But now , do I get the dev directly or ^2 each ?

Answer 10

now, we can compute the second derivative, applying the quotient rule

Answer 11

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

Answer 12

i think we will multiply by x^2 +1

Answer 13

now I substitute my formula for y', so we get:

Answer 14

\[\Large \begin{gathered}
y'' = \frac{{y'\sqrt {{x^2} + 1} - y\frac{x}{{\sqrt {{x^2} + 1} }}}}{{{x^2} + 1}} = \hfill \\
\hfill \\
= \frac{{\frac{{4y}}{{\sqrt {{x^2} + 1} }}\sqrt {{x^2} + 1} - y\frac{x}{{\sqrt {{x^2} + 1} }}}}{{{x^2} + 1}} \hfill \\
\end{gathered} \]

Answer 15

and after a simplification, I can write:
\[\Large \begin{gathered}
y'' = 4\frac{{y'\sqrt {{x^2} + 1} - y\frac{x}{{\sqrt {{x^2} + 1} }}}}{{{x^2} + 1}} = \hfill \\
\hfill \\
= 4\frac{{\frac{{4y}}{{\sqrt {{x^2} + 1} }}\sqrt {{x^2} + 1} - y\frac{x}{{\sqrt {{x^2} + 1} }}}}{{{x^2} + 1}} = \hfill \\
\hfill \\
= 4\frac{{4y\sqrt {{x^2} + 1} - yx}}{{{{\left( {{x^2} + 1} \right)}^{3/2}}}} \hfill \\
\end{gathered} \]

Answer 16

How did you get the 4 ?

Answer 17

sorry I forgotten a factor before, namely 4, and the last steps are the right steps

Answer 18

Oh, I see.

Answer 19

since the first derivative is:
\[\frac{{4y}}{{\sqrt {{x^2} + 1} }}\]

Answer 20

now, we substitute our first and second derivatives into the left side of your equation

Answer 21

so we can write:

Answer 22

By the way, can't we drive it into the proof directly ?

Answer 23

\[\Large \begin{gathered}
\left( {{x^2} + 1} \right)y'' + xy' = \hfill \\
\hfill \\
= \left( {{x^2} + 1} \right)4\frac{{4y\sqrt {{x^2} + 1} - yx}}{{{{\left( {{x^2} + 1} \right)}^{3/2}}}} + x\frac{{4y}}{{\sqrt {{x^2} + 1} }} = \hfill \\
\hfill \\
= 4\frac{{4y\sqrt {{x^2} + 1} - yx}}{{\sqrt {{x^2} + 1} }} + \frac{{4xy}}{{\sqrt {{x^2} + 1} }} = \hfill \\
\hfill \\
= \frac{{16y\sqrt {{x^2} + 1} - 4xy + 4xy}}{{\sqrt {{x^2} + 1} }} = ...? \hfill \\
\end{gathered} \]

Answer 24

we have to substitute our y' and y'' into the left side of your equation, then we have to show that left side is equal to the rigt side

Answer 25

I am doing it on paper now.

Answer 26

I got
(x^2 + 1)y'' = 16y - xy' / 4y

Answer 27

please complete my steps above

Answer 28

Ok ,

Answer 29

lol 16y

Answer 30

that's right!

Answer 31

Thanks for the help , but still wanna inquiry why did I fail to get it be like the proof ?

Answer 32

from 1st eqn
sqrt(x^2+1) = 4y/ y'

Answer 33

in those type of differential equations, by my personal experience, I can say that at each step we have to rewrite a first derivative, or second derivative using the definition of the function, as we have made, otherwise the computations will become very difficult.

Answer 34

Thanks ,I will do as you recommend. Can you help me with another one ? I will tag you in another question.

Answer 35

ok!