Is it possible to eliminate y from the equation y'=e^x/(cos(y)+1)?

  • Thread starter Thread starter adam199
  • Start date Start date
adam199
Messages
17
Reaction score
0
Title says it all. I can't seem to eliminate y and put y' in terms of x, if it is even possible at all. After messing around the closest I got was:

y'=e^x/(cos(y)+1)

I would really appreciate the help, especially if someone can tell me if it is possible at all to eliminate y from the equation, and how I can work that out.
 
Physics news on Phys.org
Hi adam199,

That is the correct result. And no, you cannot eliminate y from the expression.

However, you can take any point (x,y) = (3,4) for example, and plug it into the equation to get the instantaneous slope at that very point. Then you can use that slope to write an equation for the line tangent to y(x) at (3,4).
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top