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

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

Mathematics
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Looks like implicit differentiation, do you know how to approach this problem?
a bit. yes, it is implicit differentiation
Here, most of the work involve chain rule only

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Other answers:

differentiate each term d/dx ( ln x ) + d/dx (ln (y)
k
how can you tell by looking at the equation
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 \[2(x+2)-12(2y+3)(2)\frac{dy}{dx}=0\] \[2(x+2)-24(2y+3)\frac{dy}{dx}=0\] \[\frac{dy}{dx}=\frac{x+2}{24 y+36}\]
Because there's 2 terms together with a power, you'll have to use chain rule
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)
ok. thanks for the examples. I understand those. For the question, the book gives an answer of: y'= 1/4 ?
\[\huge D_{x}[(x+2)^2-6(2y+3)^2]=0\] \[\huge 2(x+2)-24(\frac{dy}{dx})(2x+3)=0\] 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) \[\huge \frac{dy}{dx}=\frac{1+2}{12[2(-1)+3]}\] \[\huge \frac{dy}{dx}=\frac{3}{12(-2+3)}\] \[\huge \frac{dy}{dx}=\frac{3}{12}\] \[\huge =\frac{1}{4}\]
Use chain rule when differentiating in this situation.

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