Question

Implicit differentiation
2y^3+y^2-x=0; (3,1)

Answer 1

6y^2 dy/dx+2y dy/dx=1

Answer 2

ok now solve for dy/dx and sub x=3 and y =1 so far your work looks fine

Answer 3

factor out dy/dx and isolate

Answer 4

but im a little confused on how to do that

Answer 5

easiest if you substitute the numbers in now, not later

Answer 6

true, unless the instructions are to state dy/dx

Answer 7

Hay hay I'm here

Answer 8

\[6y^2 y'+2yy'=1\]and \(y=1\) so
\[6y'+2y'=1\] etc makes it much easier to solve

Answer 9

\[\frac{dy}{dx}(6y^2+2y)=1\Rightarrow \frac{dy}{dx}=\frac{1}{6y^2+2y}\] plug in your point

Answer 10

y' is the same as dy/dx...when you're at 6y^2'+2yy'=1, you can do this:\[6y^2y'+6yy'=1\rightarrow y'(6y^2+6y)=1\rightarrow y'=1/(6y^2+6y)\]

Answer 11

1/(6y^2+2y)....sorry

Answer 12

so \[\frac{dy}{dx}(3,1)= \frac{1}{6(1)^2+2(1)}= \cdots\]

Answer 13

1/8