Homework Help: Quick Tangent Line Approximation (Derivatives)

1. Nov 9, 2008

asdfsystema

1. The problem statement, all variables and given/known data
The tangent line through given points (pi/4,pi/4) m=1

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

3. 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 :)
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Nov 9, 2008

Dick

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.

3. Nov 9, 2008

asdfsystema

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 ?

4. Nov 9, 2008

Dick

That's what I got.

5. Nov 9, 2008

Dick

Oops. You accidentally wrote sec^2(x) on the left side instead of sec^2(y), right?

6. Nov 9, 2008

asdfsystema

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?

7. Nov 9, 2008

Dick

Put x=y=pi/4 and solve for dy/dx. Or just solve for dy/dx then put x=y=pi/4. Your choice.

8. Nov 9, 2008

asdfsystema

Never mind. I found the answer. thanks again for the help :)

Last edited: Nov 9, 2008