# Having a hard time with 2 implicit equations

1. Sep 8, 2004

### Braumin

First off, I must say I truly enjoy these forums, though I tend to shy away from Calculus. Calculus and I just don't seem to get along, which is strange since I really enjoy other forms of math.

On to the questions. Note, these 3 are the ones I just can not seem to work through on an assignment for my 3rd Semester in a technology course. I may be working in the right direction here, but I don't feel like I am. Any help would be appreciated.

Question 1, using implicit differentiation only

x[ln(y)]=sin(x+y)

and the other

x[e^y]=tan^-1(xy)

I have tried several attempts at solving these but I seem to get stuck with dy/dx on both sides and no apparent way to solve for them. Any help, even if only the starting step or two, would be most appreciated!

2. Sep 9, 2004

### Muzza

If you isolate all terms containing dy/dx on one side, you can then factor out dy/dx and solve for it (this works for the first problem, probably for the second one too). A made-up example:

sin(x + y) + x * ln(y) * dy/dx = x^2 + cos(x - y) * dy/dx

<=>

x * ln(y) * dy/dx - cos(x - y) * dy/dx = x^2 - sin(x + y)

<=>

dy/dx * ( x * ln(y) - cos(x - y) ) = x^2 - sin(x + y)

<=>

dy/dx = (x^2 - sin(x + y)) / (x * ln(y) - cos(x - y))

3. Sep 9, 2004

### HallsofIvy

Staff Emeritus
There's nothing wrong with having dy/dx on both sides. Since differentiation is a linear operator it is always easy to solve for dy/dx.

(xln(y))'= ln(y)+ (x/y) dy/dx and sin(x+y)= cos(x+y)(1+ dy/dx) so

ln(y)+ (x/y) dy/dx= cos(x+y)+ cos(x+y)dy/dx
Now collect the terms involving dy/dx on the left:

((x/y)- cos(x+y))dy/dx= cos(x+y)- ln(y) and solve for dy/dx:

dy/dx= (cos(x+y)- ln(y))/((x/y)- cos(x+y))

(Did you post this under "differential equations" because you don't like calculus?)

Last edited: Sep 9, 2004