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Im referring to implicit differentiation...

spose you had y^2; then the derivaitvie is 2y y'

implicit is just the chain rule on steriods

tell me, what is the derivative of say; x^3?

\(\ \Large 3x, \text{right?} \)

no try it again, this time without the typo and ill tell you why your wrong again :)

\(\ \Large 3x^2 \text{ ?} \)

Hmmmmm......

Just x'? That's all??

x^3 derives to 3x^2 x'
now, when we define it as "with respect to x"; then x'=1

\(\ \large 1 \times y' \text{...? } \)

correct

what is the derivative of 3y^2 seeing that you know nothing about the rule that defines y

\(\ \text{Well, } 6y\times y' \text{ ? } \)

3*2[y(?)]*y'(?) hence .... 6y y'

I see.

sure

\(\ \Huge \text{THANKS! So, it is: } \)

as long as my internet connection doesnt conk out on me

power rule, constant rule, and the obiquitous chain rule ...

So like 2x/16? Doesn't the 16 become 0?

\[\frac{1}{16}x^2=\frac{x^2}{16}\]

the 16 doesnt become 0

I see now

So for the y, Ill take a shot at it:
\(\ \Large -\frac{2yy'}{9} ? \)

and so far, so good

I don't think Ive ever heard that one before "to keep things kosher..." ;)

So now I simplify? \(\ \Large \frac{2xx'}{16} \rightarrow \frac{x}{8} ? \)

As for the y, Im stuck now.

well, we know that 1'=0 for the constant, so lets write up the derivative that we have so far

Btw, what allows us to say x'=1?

im sure that somewhere in the instructions its asking you to find dy/dx

Yea, notspecifically in this set of problems, but in a previous set it did

so, write up the derivate that we have so far, then the rest should become clear

\(\ \Large \frac{x}{8} - \frac{2yy'}{9} = 0 \)

and the given point tells us the values for x and y, so plug those in as well

What about the y'? Do I need to isolate it??

yes, but i was just pointing out that y' was going to be the only unknown to solve for :)

and that y' IS the slope of thhe tangent line at the given point

Okay. :)
SO I substituted the point in, and I get y'=\(\ -\frac{41}{72} \)

Slope-Point Formula!
O
R
Or as it is more commonly known as, Point-Slope Form

correct, and dbl chk that y'
-5, 9/4
-5/8 - y'/4 = 0 --> y' = -4(5/8)
right?

Isnt it -9/4? Not y'/4? What happened to that y?

2/9 * 9/4 = 1/2 so, -2(5/8) = -10/8

9/4

\[y=\frac94\]\[\frac29yy'=\frac29\frac94y'=\frac12y'\]

Okay, that makes sense. That slope is correct btw per the book!

:)

my internet connection is toast, so ill have to say gnite :) good luck

:D