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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'???
 one year ago
 one year ago
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'???
 one year ago
 one year ago

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Study23Best ResponseYou've already chosen the best response.1
Im referring to implicit differentiation...
 one year ago

amistre64Best 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
 one year ago

amistre64Best ResponseYou've already chosen the best response.5
spose you had y^2; then the derivaitvie is 2y y'
 one year ago

Study23Best 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'?
 one year ago

amistre64Best ResponseYou've already chosen the best response.5
implicit is just the chain rule on steriods
 one year ago

amistre64Best ResponseYou've already chosen the best response.5
tell me, what is the derivative of say; x^3?
 one year ago

Study23Best ResponseYou've already chosen the best response.1
\(\ \Large 3x, \text{right?} \)
 one year ago

amistre64Best 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 :)
 one year ago

Study23Best ResponseYou've already chosen the best response.1
\(\ \Large 3x^2 \text{ ?} \)
 one year ago

amistre64Best 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
 one year ago

amistre64Best 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
 one year ago

Study23Best 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?
 one year ago

amistre64Best 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
 one year ago

Study23Best ResponseYou've already chosen the best response.1
\(\ \large 1 \times y' \text{...? } \)
 one year ago

amistre64Best 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
 one year ago

Study23Best ResponseYou've already chosen the best response.1
\(\ \text{Well, } 6y\times y' \text{ ? } \)
 one year ago

amistre64Best 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
 one year ago

amistre64Best ResponseYou've already chosen the best response.5
3*2[y(?)]*y'(?) hence .... 6y y'
 one year ago

amistre64Best 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
 one year ago

Study23Best 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?
 one year ago

Study23Best ResponseYou've already chosen the best response.1
\(\ \Huge \text{THANKS! So, it is: } \)
 one year ago

Study23Best ResponseYou've already chosen the best response.1
\(\ \Large \frac{x^2}{16}\frac{y^2}{9} = 1. \text{Find the tangent line to the curve, }\) \(\ \Large \text{ To the point} (5, \frac{9}{4}. ) \)
 one year ago

amistre64Best ResponseYou've already chosen the best response.5
as long as my internet connection doesnt conk out on me
 one year ago

amistre64Best ResponseYou've already chosen the best response.5
power rule, constant rule, and the obiquitous chain rule ...
 one year ago

Study23Best ResponseYou've already chosen the best response.1
So like 2x/16? Doesn't the 16 become 0?
 one year ago

amistre64Best ResponseYou've already chosen the best response.5
\[\frac{1}{16}x^2=\frac{x^2}{16}\]
 one year ago

amistre64Best ResponseYou've already chosen the best response.5
the 16 doesnt become 0
 one year ago

Study23Best ResponseYou've already chosen the best response.1
So for the y, Ill take a shot at it: \(\ \Large \frac{2yy'}{9} ? \)
 one year ago

amistre64Best 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
 one year ago

Study23Best ResponseYou've already chosen the best response.1
I don't think Ive ever heard that one before "to keep things kosher..." ;)
 one year ago

Study23Best ResponseYou've already chosen the best response.1
So now I simplify? \(\ \Large \frac{2xx'}{16} \rightarrow \frac{x}{8} ? \)
 one year ago

Study23Best ResponseYou've already chosen the best response.1
As for the y, Im stuck now.
 one year ago

amistre64Best 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
 one year ago

Study23Best ResponseYou've already chosen the best response.1
Btw, what allows us to say x'=1?
 one year ago

amistre64Best ResponseYou've already chosen the best response.5
im sure that somewhere in the instructions its asking you to find dy/dx
 one year ago

amistre64Best 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
 one year ago

Study23Best ResponseYou've already chosen the best response.1
Yea, notspecifically in this set of problems, but in a previous set it did
 one year ago

amistre64Best ResponseYou've already chosen the best response.5
so, write up the derivate that we have so far, then the rest should become clear
 one year ago

Study23Best ResponseYou've already chosen the best response.1
\(\ \Large \frac{x}{8}  \frac{2yy'}{9} = 0 \)
 one year ago

amistre64Best 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
 one year ago

Study23Best ResponseYou've already chosen the best response.1
What about the y'? Do I need to isolate it??
 one year ago

amistre64Best 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 :)
 one year ago

amistre64Best ResponseYou've already chosen the best response.5
and that y' IS the slope of thhe tangent line at the given point
 one year ago

Study23Best ResponseYou've already chosen the best response.1
Okay. :) SO I substituted the point in, and I get y'=\(\ \frac{41}{72} \)
 one year ago

amistre64Best 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?
 one year ago

Study23Best ResponseYou've already chosen the best response.1
SlopePoint Formula! O R Or as it is more commonly known as, PointSlope Form
 one year ago

amistre64Best 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?
 one year ago

Study23Best ResponseYou've already chosen the best response.1
Isnt it 9/4? Not y'/4? What happened to that y?
 one year ago

amistre64Best ResponseYou've already chosen the best response.5
2/9 * 9/4 = 1/2 so, 2(5/8) = 10/8
 one year ago

amistre64Best ResponseYou've already chosen the best response.5
\[y=\frac94\]\[\frac29yy'=\frac29\frac94y'=\frac12y'\]
 one year ago

Study23Best ResponseYou've already chosen the best response.1
Okay, that makes sense. That slope is correct btw per the book!
 one year ago

Study23Best ResponseYou've already chosen the best response.1
Okay, @amistre64 It looks good now! Thank you for your help the past hour! \(\ \Large \text{Words cannot express my utmost appreciation for your help!} \) \(\ \Large \text{Thank You, THANK you, THANK YOU SO MUCH!!!!! } \)
 one year ago

amistre64Best ResponseYou've already chosen the best response.5
my internet connection is toast, so ill have to say gnite :) good luck
 one year ago
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