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Study23
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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'???
 2 years ago
 2 years ago
Study23 Group Title
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'???
 2 years ago
 2 years ago

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Study23 Group TitleBest ResponseYou've already chosen the best response.1
Im referring to implicit differentiation...
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.5
the derivative of y is y' either in implicit or explicit; the rules dont change because of the name of it
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.5
spose you had y^2; then the derivaitvie is 2y y'
 2 years ago

Study23 Group TitleBest ResponseYou've already chosen the best response.1
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'?
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.5
implicit is just the chain rule on steriods
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.5
tell me, what is the derivative of say; x^3?
 2 years ago

Study23 Group TitleBest ResponseYou've already chosen the best response.1
\(\ \Large 3x, \text{right?} \)
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.5
no try it again, this time without the typo and ill tell you why your wrong again :)
 2 years ago

Study23 Group TitleBest ResponseYou've already chosen the best response.1
\(\ \Large 3x^2 \text{ ?} \)
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.5
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
 2 years ago

Study23 Group TitleBest ResponseYou've already chosen the best response.1
Hmmmmm......
 2 years ago

Study23 Group TitleBest ResponseYou've already chosen the best response.1
Just x'? That's all??
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.5
x^3 derives to 3x^2 x' now, when we define it as "with respect to x"; then x'=1
 2 years ago

Study23 Group TitleBest ResponseYou've already chosen the best response.1
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?
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.5
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
 2 years ago

Study23 Group TitleBest ResponseYou've already chosen the best response.1
\(\ \large 1 \times y' \text{...? } \)
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.5
what is the derivative of 3y^2 seeing that you know nothing about the rule that defines y
 2 years ago

Study23 Group TitleBest ResponseYou've already chosen the best response.1
\(\ \text{Well, } 6y\times y' \text{ ? } \)
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.5
think of all the variables as functions of some unknown variable 3 [y(?)]^2 and use the power and the chain rule
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.5
3*2[y(?)]*y'(?) hence .... 6y y'
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.5
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
 2 years ago

Study23 Group TitleBest ResponseYou've already chosen the best response.1
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?
 2 years ago

Study23 Group TitleBest ResponseYou've already chosen the best response.1
\(\ \Huge \text{THANKS! So, it is: } \)
 2 years ago

Study23 Group TitleBest ResponseYou've already chosen the best response.1
\(\ \Large \frac{x^2}{16}\frac{y^2}{9} = 1. \text{Find the tangent line to the curve, }\) \(\ \Large \text{ To the point} (5, \frac{9}{4}. ) \)
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.5
as long as my internet connection doesnt conk out on me
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.5
power rule, constant rule, and the obiquitous chain rule ...
 2 years ago

Study23 Group TitleBest ResponseYou've already chosen the best response.1
So like 2x/16? Doesn't the 16 become 0?
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.5
\[\frac{1}{16}x^2=\frac{x^2}{16}\]
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.5
the 16 doesnt become 0
 2 years ago

Study23 Group TitleBest ResponseYou've already chosen the best response.1
So for the y, Ill take a shot at it: \(\ \Large \frac{2yy'}{9} ? \)
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.5
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
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.5
and so far, so good
 2 years ago

Study23 Group TitleBest ResponseYou've already chosen the best response.1
I don't think Ive ever heard that one before "to keep things kosher..." ;)
 2 years ago

Study23 Group TitleBest ResponseYou've already chosen the best response.1
So now I simplify? \(\ \Large \frac{2xx'}{16} \rightarrow \frac{x}{8} ? \)
 2 years ago

Study23 Group TitleBest ResponseYou've already chosen the best response.1
As for the y, Im stuck now.
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.5
well, we know that 1'=0 for the constant, so lets write up the derivative that we have so far
 2 years ago

Study23 Group TitleBest ResponseYou've already chosen the best response.1
Btw, what allows us to say x'=1?
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.5
im sure that somewhere in the instructions its asking you to find dy/dx
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.5
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
 2 years ago

Study23 Group TitleBest ResponseYou've already chosen the best response.1
Yea, notspecifically in this set of problems, but in a previous set it did
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.5
so, write up the derivate that we have so far, then the rest should become clear
 2 years ago

Study23 Group TitleBest ResponseYou've already chosen the best response.1
\(\ \Large \frac{x}{8}  \frac{2yy'}{9} = 0 \)
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.5
and the given point tells us the values for x and y, so plug those in as well
 2 years ago

Study23 Group TitleBest ResponseYou've already chosen the best response.1
What about the y'? Do I need to isolate it??
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.5
yes, but i was just pointing out that y' was going to be the only unknown to solve for :)
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.5
and that y' IS the slope of thhe tangent line at the given point
 2 years ago

Study23 Group TitleBest ResponseYou've already chosen the best response.1
Okay. :) SO I substituted the point in, and I get y'=\(\ \frac{41}{72} \)
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.5
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?
 2 years ago

Study23 Group TitleBest ResponseYou've already chosen the best response.1
SlopePoint Formula! O R Or as it is more commonly known as, PointSlope Form
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.5
correct, and dbl chk that y' 5, 9/4 5/8  y'/4 = 0 > y' = 4(5/8) right?
 2 years ago

Study23 Group TitleBest ResponseYou've already chosen the best response.1
Isnt it 9/4? Not y'/4? What happened to that y?
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.5
2/9 * 9/4 = 1/2 so, 2(5/8) = 10/8
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.5
\[y=\frac94\]\[\frac29yy'=\frac29\frac94y'=\frac12y'\]
 2 years ago

Study23 Group TitleBest ResponseYou've already chosen the best response.1
Okay, that makes sense. That slope is correct btw per the book!
 2 years ago

Study23 Group TitleBest ResponseYou've already chosen the best response.1
Okay, @amistre64 It looks good now! Thank you for your help the past hour! \(\ \Large \text{Words cannot express my utmost appreciation for your help!} \) \(\ \Large \text{Thank You, THANK you, THANK YOU SO MUCH!!!!! } \)
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.5
my internet connection is toast, so ill have to say gnite :) good luck
 2 years ago
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