# Derivative of Trig functions problem.

1. Mar 2, 2013

### lionely

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

If x = asecθ, y =btanθ show that

dy/dx = (b/a) cosecθ and d2y/dx2 = (-b/a2)cot^3θ

The attempt at a solution

I got the 1st part

dy/dx = (dy/dθ) * (dθ/dx) = bsec^2θ x 1/(secθtanθ)= (b/a) cosecθ

Now I tried differentiating a 2nd time and I don't get the answer they want me to get. I
even tried finding d^2y/dθ and dx^2/dθ

then used the chain rule still didn't get it.

2. Mar 2, 2013

### Dick

How did you use the chain rule? d/dx=(dθ/dx)*(d/dθ). Apply that to the first derivative.

3. Mar 2, 2013

### lionely

I used the chain rule to get (b/a) cosecθ

but I can't get the 2nd derivative.

4. Mar 2, 2013

### Dick

Use it again to get the second derivative.

5. Mar 2, 2013

### lionely

Oh just do dy/dx times dy/dx?

6. Mar 2, 2013

### Dick

No. You do d/dx(dy/dx). Take the expression you got for dy/dx and find d/dx of it again. The same thing you did to get the first derivative from y.

7. Mar 2, 2013

### iRaid

$$\frac{d(\frac{dy}{dx})}{dx}$$

Maybe that makes it easier to see than Dick wrote it.

8. Mar 2, 2013

### lionely

But if I do that I get (d)/(dx)((b csc(x))/a) = -(b cot(x) csc(x))/a

9. Mar 2, 2013

### Dick

It's not csc(x). It's csc(θ). Look at it this way. Suppose you were given (b/a) cosecθ (the first derviative) and wanted to find d/dx. That's the same sort of exercise you did to get the first part.

10. Mar 2, 2013

### lionely

oh it's d(dydx)/dx = d(dy/dx)/dθ * dθ/dx? This should give me the answer right?

11. Mar 2, 2013

### Dick

Yes, that should give you the right answer.