Question

how do I find dy/dx at:
(x+2)^2-6(2y+3)^2=3, (1, -1)

Answer 1

Looks like implicit differentiation, do you know how to approach this problem?

Answer 2

a bit. yes, it is implicit differentiation

Answer 3

Here, most of the work involve chain rule only

Answer 4

differentiate each term d/dx ( ln x ) + d/dx (ln (y)

Answer 5

k

Answer 6

how can you tell by looking at the equation

Answer 7

Differentiate (x+2)^2 you'll get 2(x+2)

Differentiate -6(2y+3)^2 you'll get -12(2y+3)(2)(dy/dx)
The 2 comes from chain rule, and dy/dx is differentiating with respect to x .

Differentiate 3 is 0

Then you'll get

\[2(x+2)-12(2y+3)(2)\frac{dy}{dx}=0\]
\[2(x+2)-24(2y+3)\frac{dy}{dx}=0\]
\[\frac{dy}{dx}=\frac{x+2}{24 y+36}\]

Answer 8

Because there's 2 terms together with a power, you'll have to use chain rule

Answer 9

Examples,

1) Differentiate
3(x+6)^2
dy/dx=6(x+6)

2) Differentiate
3(4x+2)^2
dy/dx=6(4x+2)(4)
dy/dx=24(4x+2)

Answer 10

ok. thanks for the examples. I understand those. For the question, the book gives an answer of: y'= 1/4 ?

Answer 11

\[\huge D_{x}[(x+2)^2-6(2y+3)^2]=0\]
\[\huge 2(x+2)-24(\frac{dy}{dx})(2x+3)=0\]
Divide both sides by 2.
\[\huge (x+2)-12(\frac{dy}{dx})(2x+3)=0\]
Make dy/dx the subject.
\[\huge \frac{dy}{dx}12(2x+3)=(x+2)\]
\[\huge \frac{dy}{dx}=\frac{x+2}{12(2x+3)}\]
At (1, -1)
\[\huge \frac{dy}{dx}=\frac{1+2}{12[2(-1)+3]}\]
\[\huge \frac{dy}{dx}=\frac{3}{12(-2+3)}\]
\[\huge \frac{dy}{dx}=\frac{3}{12}\]
\[\huge =\frac{1}{4}\]

Answer 12

Use chain rule when differentiating in this situation.