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Study23
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'???
Im referring to implicit differentiation...
the derivative of y is y' either in implicit or explicit; the rules dont change because of the name of it
spose you had y^2; then the derivaitvie is 2y y'
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'?
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 :)
\(\ \Large 3x^2 \text{ ?} \)
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
x^3 derives to 3x^2 x' now, when we define it as "with respect to x"; then x'=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?
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
\(\ \large 1 \times y' \text{...? } \)
what is the derivative of 3y^2 seeing that you know nothing about the rule that defines y
\(\ \text{Well, } 6y\times y' \text{ ? } \)
think of all the variables as functions of some unknown variable 3 [y(?)]^2 and use the power and the chain rule
3*2[y(?)]*y'(?) hence .... 6y y'
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
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?
\(\ \Huge \text{THANKS! So, it is: } \)
\(\ \Large \frac{x^2}{16}-\frac{y^2}{9} = 1. \text{Find the tangent line to the curve, }\) \(\ \Large \text{ To the point} (-5, \frac{9}{4}. ) \)
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
So for the y, Ill take a shot at it: \(\ \Large -\frac{2yy'}{9} ? \)
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
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
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
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'=\(\ -\frac{41}{72} \)
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?
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
\[y=\frac94\]\[\frac29yy'=\frac29\frac94y'=\frac12y'\]
Okay, that makes sense. That slope is correct btw per the book!
Okay, @amistre64 It looks good now! Thank you for your help the past hour! \(\ \Large \text{Words cannot express my utmost appreciation for your help!} \) \(\ \Large \text{Thank You, THANK you, THANK YOU SO MUCH!!!!! } \)
my internet connection is toast, so ill have to say gnite :) good luck