y"

[thelazyman]

Hi, I have two y" questions.

1. y= x/root x+1, y' 1(x+1)^1/2 - x (1/2(x+1)^-1/2 / ((x+1)^1/2)^2

y'= (x+1)^1/2x(x +1) -1/2 / (x+1) ^ 3/2, y'= 1/2x + 1/ (X+1) ^ 3/2

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

the answer is -x-x/ 4root (x+1)^5


2. x^2 - 4y^2 = 4, 2x - 8ydy/dx= 0 then 2x/8y = dy/dx

then y"= 1(4y) - x 4dy/dx/ 4y^2


got confused their, the answer is y"= 1/4y^3.


Please help.


Thanks

 

[Unco]

1. y= x/root x+1, y' 1(x+1)^1/2 - x (1/2(x+1)^-1/2 / ((x+1)^1/2)^2

I can follow this much.

\(\displaystyle \L \mbox{ y' = \frac{(x+1)^{\frac{1}{2}} - \frac{1}{2}x(x+1)^{-\frac{1}{2}}}{x+1}}\)

Simplify:

\(\displaystyle \L \mbox{ y' = (x+1)^{-\frac{1}{2}} - \frac{1}{2}x(x+1)^{-\frac{3}{2}}\)

So

\(\displaystyle \L \begin{align*}
\mbox{ y'' }&=\mbox{ -\frac{1}{2}(x+1)^{-\frac{3}{2}} + \frac{3}{4}x(x+1)^{-\frac{5}{2}} - \frac{1}{2}(x+1)^{-\frac{3}{2}}}\\
\\
\\
\\
\\
\\
\\

&= \mbox{-(x+1)^{-\frac{3}{2}} + \frac{3}{4}x(x+1)^{-\frac{5}{2}}}
\end{align*}\)

Factorise

\(\displaystyle \L
\begin{align*}
\mbox{ y'' }&=\mbox{ (x+1)^{-\frac{5}{2}}\left(-(x+1) + \frac{3}{4}x\right)}\\
\\
\\
\\
\\
\\
\\

&= \mbox{\frac{-\frac{1}{4}x - 1}{(x+1)^{\frac{5}{2}}}}\\
\\
\\
\\
\\
\\
\\

&= \mbox{\frac{-x - 4}{4\sqrt{(x+1)^{5} }}}\\
\end{align*}\)

2. x^2 - 4y^2 = 4, 2x - 8ydy/dx= 0 then 2x/8y = dy/dx

then y"= 1(4y) - x 4dy/dx/ 4y^2

got confused their, the answer is y"= 1/4y^3.

Substitute back dy/dx = x/4y to simplify y''.

Their answer is missing a negative sign.

 

[ChaoticLlama]

1) I'm assuming the first equation is this: \(\displaystyle y = \frac{x}{{\sqrt {x + 1} }}\)

There is no simplification for this, quotient rule must be used. Remember that is by definition:
For a function \(\displaystyle y = \frac{{p(x)}}{{q(x)}}\)
It's derivative is equal to \(\displaystyle \frac{{dy}}{{dx}} = \frac{{q(x)p'(x) - p(x)q'(x)}}{{[q(x)]^2 }}\)

Therefore, continue as such...
\(\displaystyle \begin{array}{l}
y = \frac{x}{{\sqrt {x + 1} }} \\
\frac{{dy}}{{dx}} = \frac{{\sqrt {x + 1} *(1) - x*\frac{1}{2}(x + 1)^{\frac{{ - 1}}{2}} }}{{x + 1}} \\
\frac{{dy}}{{dx}} = \frac{{\sqrt {x + 1} - \frac{x}{{2\sqrt {x + 1} }}}}{{x + 1}} \\
\end{array}\)

Does your first derivative look like that? The second derivative is messy no matter which way you look at it. Post reply if you are having your difficulties in your second differentiation.

2) x² - 4y² = 4
This one is in my opinion, much easier considering that there is no quotient to deal with.

First derivative:
\(\displaystyle \begin{array}{c}
x^2 - 4y^2 = 4 \\
2x - 8y\frac{{dy}}{{dx}} = 0 \\
8y\frac{{dy}}{{dx}} = 2x \\
\frac{{dy}}{{dx}} = \frac{x}{{4y}} \\
\end{array}\)

Second Derivative:
\(\displaystyle \begin{array}{l}
\frac{{dy}}{{dx}} = \frac{x}{{4y}} \\
\frac{{d^2 y}}{{dx^2 }} = \frac{{4y - 4x\frac{{dy}}{{dx}}}}{{(4y)^2 }} \\
\frac{{d^2 y}}{{dx^2 }} = \frac{{4y - 4x\left( {\frac{x}{{4y}}} \right)}}{{(4y)^2 }} \\
\frac{{d^2 y}}{{dx^2 }} = \frac{{4y - \left( {\frac{{x^2 }}{y}} \right)}}{{(4y)^2 }} \\
\end{array}\)

 

[Unco]

Thelazyman: I neglected to mention, through forgetfulness, for you to use the original equation to further simplify y''.

 

[thelazyman]

Just wondering UNCO how do you get the math in that format.

 

[stapel]

Just wondering UNCO how do you get the math in that format.

To learn how to format using LaTeX, please follow the links in the "Forum Help" pull-down menu at the very top of the page.

Thank you.

Eliz.

 

[Unco]

Just wondering UNCO how do you get the math in that format.

Click "quote" next to a person's message. Latex (what's between the [ tex ] tags) looks daunting at first but play around with it and you'll get the hang of it soon enough.

Note that some people use a tex editor which spits out ugly code, so watch for that.