# Point on a curve question

1. Feb 24, 2008

### Flappy

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

Find the point, restricting $$x\in$$(-pi/2, 0), along the curve

f(x) = $$\frac{e^x}{cos(x)}$$

where the slope of the tangent line to f has zero slope.

3. The attempt at a solution
I found the derivative of f(x) and got:

$$\frac{e^x(cosx - sinx)}{cos^2(x)}$$

Where would I go from here to find the point though?

2. Feb 24, 2008

### Mystic998

How does the derivative relate to the slope of a tangent line to the curve?

Edit: Incidentally, the expression you've given as the derivative is incorrect.

Last edited: Feb 24, 2008
3. Feb 24, 2008

### sutupidmath

well remember that the slope of the tangent line at a point in a curve is merely the derivative of that function at that point. so since you found teh derivative of the function f(x) that means that you have found the slope of that function at any point. But y ou are interested only when that slope is zero . so just equal your rezult to zero and solve for x, if you can.
$$\frac{e^x(cosx + sinx)}{cos^2(x)}=0$$

Last edited: Feb 24, 2008
4. Feb 24, 2008

### sutupidmath

this looks to me like it does not have an explicit solution though!!!

5. Feb 24, 2008

### rocomath

$$f'(x)=\frac{e^x(\cos x+\sin x)}{\cos^{2}x}$$