Question

Take the second derivative of
y= sqrt(x^2+1)

Answer 1

take the first derivative

Answer 2

Yes, and I got
y'= x(x^2+1)^-1/2

Answer 3

then take the derivative of that

Answer 4

Do I use the product rule of second derivative?

Answer 5

how would you normally solve that?

Answer 6

I got
y " = -x^2(x^2+1)^-3/2

Answer 7

Hello?

Answer 8

@nincompoop ??

Answer 9

Alternative : \[y = \sqrt{x^2+1} \implies y^2 = x^2+1\]
implicitly differentiating both sides gives
\[2yy' = 2x \implies yy' = x\]

differentiate again implicitly and solve \(y'\)

Answer 10

hint:
you have to apply the derivative of a quotient between two functions, namely:
\[\Large y'' = \frac{{1 \cdot \sqrt {{x^2} + 1} - x \cdot \frac{x}{{\sqrt {{x^2} + 1} }}}}{{{x^2} + 1}} = ...?\]

Answer 11

since, the first derivative, is:
\[\Large y' = \frac{x}{{\sqrt {{x^2} + 1} }}\]

Answer 12

ye you can use quotient rule for that, but you can also use product rule with with chain rule

Answer 13

it depends on how you can strategize and make it simpler for you to solve

Answer 14

I tried usinf the product rule and chain rule, but I kept on getting
y '' = -x^2(x^2+1)^-3/2

Answer 15

write the product rule for me

Answer 16

Easy to make mistakes with those messy expressions. Just take a deep breath and double check ur work

Answer 17

laughing inside my head and physically smiling

Answer 18

are you sure? @esam2

Answer 19

I got this for using the product rule and chain rule:
\[y''= (x^2+1)^{-1/2}+x(-1/2(x^2+1)^{-3/2}(2x)\]

Answer 20

@rational @Michele_Laino

Answer 21

using the quotient rule, I got this:
\[\Large \begin{gathered}
y'' = \frac{{1 \cdot \sqrt {{x^2} + 1} - x \cdot \frac{x}{{\sqrt {{x^2} + 1} }}}}{{{x^2} + 1}} = \frac{{{x^2} + 1 - {x^2}}}{{{{\left( {{x^2} + 1} \right)}^{3/2}}}} = \hfill \\
\hfill \\
= \frac{1}{{{{\left( {{x^2} + 1} \right)}^{3/2}}}} \hfill \\
\end{gathered} \]

Answer 22

That is the right answer, I'm just wondering how to do it by using the product rule...

Answer 23

then, we have to write the first derivative, as below:
\[y' = \frac{x}{{\sqrt {{x^2} + 1} }} = x \cdot \frac{1}{{\sqrt {{x^2} + 1} }} = x{\left( {{x^2} + 1} \right)^{ - 1/2}}\]

Answer 24

Yes, I got that part

Answer 25

so, we have:
\[y'' = 1 \cdot {\left( {{x^2} + 1} \right)^{ - 1/2}} + x\left( { - \frac{1}{2}{{\left( {{x^2} + 1} \right)}^{ - 3/2}}2x} \right) = ...?\]

Answer 26

\[\begin{gathered}
y'' = 1 \cdot {\left( {{x^2} + 1} \right)^{ - 1/2}} + x\left( { - \frac{1}{2}{{\left( {{x^2} + 1} \right)}^{ - 3/2}}2x} \right) = \hfill \\
= \frac{1}{{\sqrt {{x^2} + 1} }} - \frac{{{x^2}}}{{{{\left( {{x^2} + 1} \right)}^{3/2}}}} = \frac{{{x^2} + 1 - {x^2}}}{{{{\left( {{x^2} + 1} \right)}^{3/2}}}} \hfill \\
\end{gathered} \]

Answer 27

How did you get it like that?

Answer 28

I got that, using the properties of the real numbers

Answer 29

What is the properties of the real numbers?

Answer 30

for example, what is the least common multiple between:
\[{\sqrt {{x^2} + 1} }\]
and
\[{{{\left( {{x^2} + 1} \right)}^{3/2}}}\]?

Answer 31

I didn't quite understand how to get this part:\[\frac{ x^2+1+x^2 }{ (x^2+1)^{3/2} }\]

Answer 32

sqrt(x^2+1)?

Answer 33

no, it is:
\[{{{\left( {{x^2} + 1} \right)}^{3/2}}}\]

Answer 34

Oh..

Answer 35

so, we can write:
\[\large \frac{1}{{\sqrt {{x^2} + 1} }} - \frac{{{x^2}}}{{{{\left( {{x^2} + 1} \right)}^{3/2}}}} = \frac{{1 \cdot \left( {{x^2} + 1} \right) - {x^2} \cdot 1}}{{{{\left( {{x^2} + 1} \right)}^{3/2}}}}\]

Answer 36

Oh I see!

Answer 37

I think I got it from here. Thanks for all your help! :)

Answer 38

Thanks! :)