Quick Tangent Line Approximation (Derivatives)

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SUMMARY

The discussion focuses on the process of finding the derivative of the function defined by the equation y = cos(y)cos(x)/sin(x)sin(y) at the point (π/4, π/4). Participants confirm the use of the quotient rule and product rule for differentiation, emphasizing the importance of implicit differentiation. The correct approach involves rearranging the equation to y*tan(y) = tan(x) and applying the product rule, leading to the expression y*sec²(y)*dy/dx + tan(y)*dy/dx = sec²(x). The final step is substituting x = y = π/4 to solve for dy/dx.

PREREQUISITES
  • Understanding of implicit differentiation
  • Familiarity with the quotient rule and product rule in calculus
  • Knowledge of trigonometric functions and their derivatives
  • Ability to apply the chain rule in differentiation
NEXT STEPS
  • Study implicit differentiation techniques in calculus
  • Review the application of the product rule and quotient rule
  • Learn about the derivatives of trigonometric functions
  • Practice solving differential equations involving trigonometric identities
USEFUL FOR

Students studying calculus, particularly those focusing on derivatives and implicit differentiation, as well as educators looking for examples of solving complex differentiation problems.

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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