Question

find second derivative of y^(1/2)+xy

Answer 1

please explain how you arrived at your answer

Answer 2

i see a power rule, a product rule, and a chain rule ... assuming this is implicits

Answer 3

yes

Answer 4

im just having a little trouble with a few of my homework problems

Answer 5

the main issue with implicits is that you are only used to working with y; but forget that x and y mean anything and just consider that operations that relate to them.

tell me, how would you think a power rule for y^4 looks like?

Answer 6

4y^3y'

Answer 7

very good, and for this one just apply the exponent it uses ... and a product rule for xy ?

Answer 8

i know that the first derivative should be 1/2y^-(1/2)y'+y+y' right? but I can't figure out the second derivative of this

Answer 9

\[y^{1/2}+xy\]
\[\frac12y^{-1/2}~y'+(x'y+xy')\]
solve for y' cause we will need it in the second derivative; and x' wrtx = 1

\[D[\frac12y^{-1/2}~y'+y+xy']\]
\[D[\frac12y^{-1/2}~y']+D[y]+D[xy']\]
\[\frac12D[y^{-1/2}~y']+D[y]+D[xy']\\power/product~~~~~~~~~~~~~~~product\]

Answer 10

\[\frac12D[y^{-1/2}~y']+D[y]+D[xy']\]
\[\frac12D[y^{-1/2}~y']+y'+x'y'+xy''\]
\[\frac12(-\frac12y^{-3/2}~y'y'+y^{-1/2}~y'')+y'+y'+xy''\]
\[-\frac14y^{-3/2}~(y')^2+\frac12y^{-1/2}~y''+y'+y'+xy''\]

solve for y''

Answer 11

\[-\frac14y^{-3/2}~(y')^2+\frac12y^{-1/2}~y''+y'+y'+xy''\]
\[-\frac14y^{-3/2}~(y')^2+\frac12y^{-1/2}~y''+2y'+xy''\]
\[-\frac14y^{-3/2}~(y')^2+y''(\frac12y^{-1/2}+x)+2y'\]
\[y''(\frac12y^{-1/2}+x)=\frac14y^{-3/2}~(y')^2-2y'\]
\[y''=\frac{\frac14y^{-3/2}~(y')^2-2y'}{\frac12y^{-1/2}+x}\]

Answer 12

you can simplify it to your hearts content .....

Answer 13

thanks I'm going to work on this and try to get the hang of it.

Answer 14

good luck :) i find it simpler to get the form of the second derivative and then fill in the y' parts; but sometimes it might be simpler to simplify the y's first .... personal preferences and all

Answer 15

i"m just having trouble with all the negative powers and all the rational powers it makes it really confusing especially when I can't just leave them in the form they always want it positive and with the roots. For example the answer the book got was \[\frac{ 2y ^{2}(5+8x \sqrt{y}) }{ (1+2x \sqrt{y})^{3} }\]

Answer 16

lets determine a suitable expression for y'

\[\frac12y^{-1/2}~y'+(x'y+xy')\]
\[\frac12y^{-1/2}~y'+y+xy'\]
\[y'(\frac12y^{-1/2}+x)=-y\]
\[y'=-\frac{y}{\frac12y^{-1/2}+x}\]
and sort it
\[y'=-\frac{y}{\frac1{2\sqrt{y}}+x}\]
\[y'=-\frac{2y\sqrt{y}}{1+2x\sqrt{y}}\]

now, we have something to insert into the y' parts

Answer 17

lets clean up the y''
\[y''=\frac{\frac14y^{-3/2}~(y')^2-2y'}{\frac12y^{-1/2}+x}\]
\[y''=\frac{\frac{(y')^2}{4y\sqrt{y}}-2y'}{\frac1{2\sqrt{y}}+x}\]
\[y''=\frac{\frac{(y')^2}{2y}-4\sqrt{y}y'}{1+2x\sqrt{y}}\]
\[y''=\frac{(y')^2-8y\sqrt{y}y'}{2y(1+2x\sqrt{y})}\]
with any luck i mathed that out correctly

Answer 18

\[y''=\frac{(-\frac{2y\sqrt{y}}{1+2x\sqrt{y}})^2-8y\sqrt{y}(-\frac{2y\sqrt{y}}{1+2x\sqrt{y}})}{2y(1+2x\sqrt{y})}\]

\[y''=\frac{\frac{4y^3}{1+2x^2y+4x\sqrt{y}}+\frac{16y^3}{1+2x\sqrt{y}}}{2y(1+2x\sqrt{y})}\]

your right, it is a pain :)

Answer 19

lets assume i did the y' simplification correctly :)

\[y'=-\frac{2y\sqrt{y}}{1+2x\sqrt{y}}\]
\[y'=-\frac{(1+2x\sqrt{y})(2y\sqrt{y})'-....}{1+2x\sqrt{y}}\]still a pain; lets do a product instead of a quotient

Answer 20

\[y'=-(2y\sqrt{y})({1+2x\sqrt{y})^{-1}}\]
\[y'=-(2yy^{1/2})({1+2xy^{1/2})^{-1}}\]
\[y'=-2(y'y^{1/2}+\frac12yy^{-1/2})({1+2xy^{1/2})^{-1}}\]
lol .... its just messy no matter what i think of

Answer 21

\[y''=\frac{\frac{4y^3}{(1+2x\sqrt{y})^2}+\frac{16y^3}{1+2x\sqrt{y}}}{2y(1+2x\sqrt{y})}\]
\[y''=2y\frac{\frac{2y^2}{(1+2x\sqrt{y})^2}+\frac{8y^2}{1+2x\sqrt{y}}}{2y(1+2x\sqrt{y})}\]
\[y''=\frac{\frac{2y^2}{(1+2x\sqrt{y})^2}+\frac{8y^2}{1+2x\sqrt{y}}}{1+2x\sqrt{y}}\]
\[y''=\frac{\frac{2y^2}{(1+2x\sqrt{y})^2}+\frac{8y^2(1+2x\sqrt{y})}{(1+2x\sqrt{y})^2}}{1+2x\sqrt{y}}\]
\[y''=\frac{2y^2+8y^2(1+2x\sqrt{y})}{(1+2x\sqrt{y})(1+2x\sqrt{y})^2}\]
\[y''=\frac{2y^2(1+4(1+2x\sqrt{y})}{(1+2x\sqrt{y})^3}\]
looking closer

Answer 22

wow thank you so much you my good fellow have gone above and beyond the call of duty

Answer 23

lol, it was bothering me :) good luck

Answer 24

if this came up in real life; i would have left it as 2 variable y'' and y'; solving for y' would have created a solid value to apply in the y'' and we could have avoided all the senseless simplifications ....

Answer 25

most definitely it was one of the last problems on the homework anyway probably not going to be anything near that difficult on a test but i just wanted to know how to do it.