Question

Use implicit differentiation to find the slope of the tangent line to the curve 2 x^2 - 2 xy + 3 y^3 = -18 at the point (-3,-2)

Answer 1

\[ 2 x^2 - 2 xy + 3 y^3 = -18 \] start with
\[4x-2y-2xy'+9y^2y'=0\] then solve for \(y'\)

Answer 2

probably easiest to solve if you plug in \((-3,2)\) for \(x\) and \(y\)

Answer 3

do you mean (-3,-2)?

Answer 4

whatever it is, plug it in, makes the algebra mostly arithmetic

Answer 5

thank you!

Answer 6

yw

Answer 7

Use implicit differentiation to find the slope of the tangent line to the curve (y/(x-6y))=x^8-2 at the point (1,(-1/-5))

Answer 8

\[y=x^2\] We know the answer to this right? Well what if you wanted to take the derivative of this? Wouldn't it be exactly the same? \[\sqrt{y} = x\]\[y^{1/2}=x^1\]Now take the derivative but use the chain rule on everything and take absolutely no shortcuts to make our point absolutely clear.\[\frac{ 1 }{ 2}*y^{-1/2}*\frac{ dy }{ dx }=1*x^0*\frac{ dx }{ dx }\]
So what is x^0? Anything to the zero power is just 1. How about dx/dx that looks silly. But how much does x change with respect to x? You have exactly a 1 to 1 correspondence here so the whole right side is 1. Solving for dy/dx we get:\[\frac{ dy }{ dx }=2y^{1/2}\]plug in y=x^2 from before and you'll see that it all works out.

I hope this wasn't a complete waste of your time, I'm just showing you that you've always ever done implicit differentiation and shouldn't be afraid by it.

Answer 9

thank you so much!!