# Quick Tangent Line Approximation (Derivatives)

## 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 :)

## The Attempt at a Solution

Dick
Homework Helper
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 ?

Dick
Homework Helper
That's what I got.

Dick
Homework Helper
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?

Dick