Quick Tangent Line Approximation (Derivatives)

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Homework Help Overview

The discussion revolves around finding the tangent line approximation for a function defined implicitly by the equation y = cos(y)cos(x)/sin(x)sin(y) at the point (π/4, π/4). The subject area is derivatives and implicit differentiation.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of the quotient rule and product rule in differentiation, with some questioning the need for implicit differentiation. There is exploration of rearranging the function to facilitate differentiation.

Discussion Status

Participants are actively engaging with the problem, offering suggestions and corrections. Some have expressed uncertainty about their differentiation steps, while others have confirmed similar approaches. There is a collaborative effort to clarify the differentiation process and resolve any misunderstandings.

Contextual Notes

There is mention of potential confusion regarding the variables used in differentiation, as well as the need to evaluate the derivative at a specific point. Participants are navigating through the implications of implicit differentiation and the correct application of differentiation rules.

asdfsystema
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Homework Statement


The tangent line through given points (pi/4,pi/4) m=1

y= cos(y)cos(x)/sin(x)sin(y)


The Attempt at a Solution


dy/dx= d/dx[cos(y)cos(x)/sin(x)sin(y)]

First use quotient rule ?
vu'-uv'/v^2

v= sin(x)sin(y)
v'= do i need to use product rule? for product rule I got sin(x)*cos(x)+sin(x)*cos(x)
but if I do it normal way, I get cos(x)*sin(y)
u= cos(y)cos(x)
u'= -sin(x)cos(y)

Quotient Rule:
sin(x)sin(y)*-sin(x)cos(y) - cos(y)cos(x)*cos(x)*sin(y)
sin(x)sin(y)^2


I just need to know if I'm going in the right direction. I'm pretty sure I know what to do for the point slope formula since its basic algebra. But it's this beginning part that I have problems with ! Thanks in advance :)
 
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Yes, you need to use the product rule. You also need to use the chain rule. d/dx(cos(y))=-sin(y)*dy/dx. You might find it easier to rearrange your original function into y*tan(y)=tan(x). Now take d/dx of both sides. You'll get more than one dy/dx. Treat it as a unknown and solve for it. This is called implicit differentiation.
 
Thanks ! I never thought about changing it ...

I followed your suggestion and took d/dx of both sides

y*tan(y) use product rule-->
u=y u'= dy/dx
v=tan(y) v'= sec^2(y) * dy/dx ?

y*sec^2(x)*dy/dx+ tan(y)*dy/dx = sec^2(x)

Is this correct ?
 
That's what I got.
 
asdfsystema said:
Thanks ! I never thought about changing it ...

I followed your suggestion and took d/dx of both sides

y*tan(y) use product rule-->
u=y u'= dy/dx
v=tan(y) v'= sec^2(y) * dy/dx ?

y*sec^2(x)*dy/dx+ tan(y)*dy/dx = sec^2(x)

Is this correct ?

Oops. You accidentally wrote sec^2(x) on the left side instead of sec^2(y), right?
 
oops yes, sorry . too used to using x. yes i meant sec^2(y) !

how do i go about eliminating one of the dy/dx ? What are the next steps?
 
Put x=y=pi/4 and solve for dy/dx. Or just solve for dy/dx then put x=y=pi/4. Your choice.
 
Never mind. I found the answer. thanks again for the help :)
 
Last edited:

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