Question

use implicit differentiation to find y'' if xy+y-x=1

Answer 1

i have found y' to be \[\frac{ 1-y }{ x+1 }\] but when i try to calculate second derivative i get \[\frac{ 2(y-1) }{ (x+1)^2 }\] when the answer should be \[\frac{ y-1 }{ (x+1)^2}\]

Answer 2

1) apply the derivative operator d/dx to xy+y-x=1.
2) Solve the resulting equation for y '.
3) Again apply the derivative operator, this time to y '
4) Solve the resulting equation for y ''
5) (Optional) Replace y ' in your last result with the expression you obtained for it.

Answer 3

xy+y-x=1
y(x+1)=x+1
\[y=\frac{ x+1 }{ x+1 }=1,y'=0,y"=0\]

Answer 4

the answer is not 0 @surjithayer

Answer 5

@amyvincent12 : I surely appreciate your showing your own work. let me check that work quickly and then get back to you.

Answer 6

if i show you my workings @mathmale can you show me where i went wrong?

Answer 7

Yes. I'm going to look thru what you've already posted; if you have more, please post that also. Thanks. Great approach.

Answer 8

well if y' =\[\frac{ 1-y }{ x+1}\]

Answer 9

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

Answer 10

Amy: Your result for y " is correct.
And I see you're correctly applying the quotient rule to y '.
How would you now simplify your expression for y ''?

Answer 11

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

Answer 12

You are really accurate!
All you need to do now is to (1) simplify your last eqxpression and (2) replace y ' in the second derivative with (1-y) / (x+1).

Answer 13

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

Answer 14

i got this by plugging in y' i got earlier and then putting negative into brackets, the (x+1)s on top part of equation cancel which leaves with \[y''=\frac{ 2(y-1) }{ (x+1)^2 }\]

Answer 15

Altho I'm not going to check every detail, both your explanation and your actual work look appropriate to me. I'm impressed, and happy for you. Feel satisfied with this result?

Answer 16

no because if i plug this question into wolfram alpha app it tells me the answer of the question should not have 2 in numerator

Answer 17

can i use quotient rule with implicit differentiation or is that not allowed?

Answer 18

1. If our result differs from that of wolframapha.com, it's most likely due to a simple algebraic error.
2. You can definitely use the quotient rule with implicit diff'n.

My suggestion is that you re-do the problem, quickly; doing so shouldn't take long. Then compare your new work to your previous work. If this doesn't help you, I'll go through the problem again myself.

Answer 19

i have done problem 4 times and getting same answer

Answer 20

xy+y-x=1
xy'+y*1+y'-1=0
xy''+y'*1+y'+y''=0
y''(x+1)=-2y'
\[y''=-\frac{ 2 }{ x+1 }y'\]
\[y'=\frac{ 1-y }{ x+1 }\]
\[y''=\frac{ 2\left( y-1 \right) }{ \left( x+1 \right)^2 }\]
if we put y=1 in y' and y''
we get y'=0,y''=0

Answer 21

your question may be xy+y+x=1

Answer 22

no it is xy+y-x=1

Answer 23

then whatever you calculated is correct.
may i know the answer.

Answer 24

Amy: My suggestion is that you move on to other problems. If you can finish the others and still have time left over, contact me or surjithayer for additional help. You've done a very commendable job of documenting your efforts and of doing the math correctly.