Question

Help Please!! (:
(2+y)^3 + 2y = x+27
Evaluate dy/dx at the point (2,1).

Answer 1

\[(2+y)^3 + 2y = x+27\] right?

Answer 2

Yes!(:

Answer 3

it would probably be a lot easier to find \(\frac{dx}{dy}\) since this is actually written as a function of \(y\) but no matter
take the derivative via the chain rule and get
\[3(2+y)^2y'+2y'=1\]

Answer 4

then solve for \(y'\) and replace \(y\) by \(1\)

Answer 5

alternatively, replace \(y\) by \(1\) at the start
probably easiest method

Answer 6

\[3(2+y)^2y'+2y'=1\]
\[3(2+1)^2y'+2y'=1\] etc

Answer 7

so \[(2+1)^3+2(1)=x+27\]

Answer 8

no

Answer 9

you are looking for \(y'\)

Answer 10

wait nvm, my mistake, so how do I get y' ?

Answer 11

i am sorry, i confused you with my last statement, let me start again

Answer 12

you can't replace \(y\) by \(1\) at the start, you will never find \(y'\) that way
what i meant is, since you have \[3(2+y)^2y'+2y'=1\] you can either
a) solve for \(y'\) using algebra, then replace \(y\) by \(1\) or
b) replace \(y\) by \(1\) now, then solve for \(y'\)

Answer 13

So I evaluate it and get
27y' +2y'=1
29y'=1
y'= 1/29
?

Answer 14

that is what i get, yes

Answer 15

Thank you!!!!(: @satellite73

Answer 16

yw