I've been getting mixed answers from people. When do I know when I take the derivative of y to have yy' versus just y'???

- anonymous

- schrodinger

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- anonymous

Im referring to implicit differentiation...

- amistre64

the derivative of y is y' either in implicit or explicit; the rules dont change because of the name of it

- amistre64

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

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## More answers

- anonymous

That's where I am confused. How do I know I should have yy', when in other cases you should have y'? In what cases do I only have y'?

- amistre64

implicit is just the chain rule on steriods

- amistre64

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

- anonymous

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

- amistre64

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

- anonymous

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

- amistre64

thats better, but still wrong and heres why
you are so used to taking a derivative "with respect to x" that you blindly assume that when the "with respect to" is not defined that it automatically has to be "with respect to x"
since we dont know what this is in respect to, the chain rule pops out an x' until further notice

- anonymous

Hmmmmm......

- anonymous

Just x'? That's all??

- amistre64

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

- anonymous

I thought you said we cant assume it to be the derivate with respect to x? Then we can't take the derivative of 3x^2?

- amistre64

the point is, without knowing what it is derived with respect to .... the chain rule pops out a prime of the function
what is the derivative of y? seeing that you know nothing of the rule that defines y

- anonymous

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

- amistre64

correct

- amistre64

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

- anonymous

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

- amistre64

think of all the variables as functions of some unknown variable
3 [y(?)]^2 and use the power and the chain rule

- amistre64

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

- anonymous

I see.

- amistre64

when you start learning calc, they keep it simpla and only define things in terms of some known rules and it seems to put a mistaken idea in the heads about what is actually happening

- anonymous

Hmmm... Okay. Do you think you could help me with a implicit differentiation problem that I've been stuck on for like over an hour?

- amistre64

sure

- anonymous

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

- anonymous

\(\ \Large \frac{x^2}{16}-\frac{y^2}{9} = 1. \text{Find the tangent line to the curve, }\)
\(\ \Large \text{ To the point} (-5, \frac{9}{4}. ) \)

- amistre64

as long as my internet connection doesnt conk out on me

- amistre64

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

- anonymous

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

- amistre64

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

- amistre64

the 16 doesnt become 0

- anonymous

I see now

- anonymous

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

- amistre64

go ahead and pop in an x' just to keep things kosher, in the end x'=1 but its good to root that in your mindset

- amistre64

and so far, so good

- anonymous

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

- anonymous

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

- anonymous

As for the y, Im stuck now.

- amistre64

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

- anonymous

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

- amistre64

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

- amistre64

the phrase implicit even suggests that y is a function of something ... and at this stage they infer that that something is rule invloving x

- anonymous

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

- amistre64

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

- anonymous

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

- amistre64

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

- anonymous

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

- amistre64

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

- amistre64

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

- anonymous

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

- amistre64

ill trust you on the math for that one; now how do we define a line if we know th eslope and a point on the line?

- anonymous

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

- amistre64

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

- anonymous

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

- amistre64

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

- anonymous

9/4

- amistre64

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

- anonymous

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

- amistre64

:)

- anonymous

Okay, @amistre64 It looks good now! Thank you for your help the past hour!
\(\ \Large \text{Words cannot express my utmost appreciation for your help!} \)
\(\ \Large \text{Thank You, THANK you, THANK YOU SO MUCH!!!!! } \)

- amistre64

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

- anonymous

:D

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