Question

Find d2y/dx2 in terms of x and y.

x^2y^2 - 16x = 4

Answer 1

did you find \(y'\) first?

Answer 2

by 'implicit differentiation"?

Answer 3

y' would be \[\frac{2\left(2x+1\right)\sqrt{\left(x\right)^2}}{x^3\sqrt{4x+1}}\] ?

Answer 4

i see no square roots in your question

Answer 5

you posted this :\[x^2y^2 - 16x = 4\]

Answer 6

What I did was put everything in terms of y? Would that be wrong?

Answer 7

yes it would

Answer 8

So, how would I approach this problem?

Answer 9

you familiar with implicit differentiation?

Answer 10

Yes

Answer 11

use that

Answer 12

you should get \[2xy^2+2x^2yy'-16=0\] solve for \(y'\)

Answer 13

Dy/Dx = (4+16x)/(2y2x) ?

Answer 14

hmm no

Answer 15

\[2xy^2+2x^2yy'-16=0\\
xy^2+x^2yy'-4=0\\
xy^2+x^2yy'=4\\
x^2yy'=4-xy^2\]

Answer 16

so \[y'=\frac{4-xy^2}{x^2y}\]

Answer 17

How would I then solve for the 2nd derivative? Thank you

Answer 18

take the derivative of that !

Answer 19

requires the quotient rule, and is kind of messy, but totally doable

Answer 20

Thank you :). I guess my teacher enjoys the suffering of others