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anonymous
 3 years 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'???
anonymous
 3 years 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'???

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Im referring to implicit differentiation...

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0That'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
 3 years ago
Best ResponseYou've already chosen the best response.5implicit is just the chain rule on steriods

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\(\ \Large 3x, \text{right?} \)

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\(\ \Large 3x^2 \text{ ?} \)

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.5thats 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
 3 years ago
Best ResponseYou've already chosen the best response.0Just x'? That's all??

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I 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
 3 years ago
Best ResponseYou've already chosen the best response.5the 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
 3 years ago
Best ResponseYou've already chosen the best response.0\(\ \large 1 \times y' \text{...? } \)

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\(\ \text{Well, } 6y\times y' \text{ ? } \)

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

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.53*2[y(?)]*y'(?) hence .... 6y y'

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.5when 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
 3 years ago
Best ResponseYou've already chosen the best response.0Hmmm... Okay. Do you think you could help me with a implicit differentiation problem that I've been stuck on for like over an hour?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\(\ \Huge \text{THANKS! So, it is: } \)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\(\ \Large \frac{x^2}{16}\frac{y^2}{9} = 1. \text{Find the tangent line to the curve, }\) \(\ \Large \text{ To the point} (5, \frac{9}{4}. ) \)

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So like 2x/16? Doesn't the 16 become 0?

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

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.5the 16 doesnt become 0

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So for the y, Ill take a shot at it: \(\ \Large \frac{2yy'}{9} ? \)

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.5go 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I don't think Ive ever heard that one before "to keep things kosher..." ;)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So now I simplify? \(\ \Large \frac{2xx'}{16} \rightarrow \frac{x}{8} ? \)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0As for the y, Im stuck now.

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Btw, what allows us to say x'=1?

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

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.5the 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
 3 years ago
Best ResponseYou've already chosen the best response.0Yea, notspecifically in this set of problems, but in a previous set it did

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\(\ \Large \frac{x}{8}  \frac{2yy'}{9} = 0 \)

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0What about the y'? Do I need to isolate it??

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Okay. :) SO I substituted the point in, and I get y'=\(\ \frac{41}{72} \)

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.5ill 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
 3 years ago
Best ResponseYou've already chosen the best response.0SlopePoint Formula! O R Or as it is more commonly known as, PointSlope Form

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Isnt it 9/4? Not y'/4? What happened to that y?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.52/9 * 9/4 = 1/2 so, 2(5/8) = 10/8

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Okay, that makes sense. That slope is correct btw per the book!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Okay, @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
 3 years ago
Best ResponseYou've already chosen the best response.5my internet connection is toast, so ill have to say gnite :) good luck
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