# Derivative help

1. Feb 13, 2009

### ldbaseball16

derivative help!!!!!!!

1. The problem statement, all variables and given/known data
find f'(x) sinhx/coshx

2. Relevant equations
this is what i got but im unsure if its wrong could anyone help me pleasee

3. The attempt at a solution
coshx(sinhx)-(coshx)coshx/coshx......... is this right if not help please

2. Feb 13, 2009

### Dick

Re: derivative help!!!!!!!

How are you getting that? Write down the quotient rule for the derivative of f(x)/g(x) and check the parts.

3. Feb 13, 2009

### ldbaseball16

Re: derivative help!!!!!!!

ok i took a look at the quotient rule and was completely wrong how does this look.....

coshx(coshx)-sinhx(sinhx)/(coshx)^2.............. (Coshx)^2-(Sinhx)^2/(Coshx)^2..........the (coshx)^2 cancel out and you are left with -(sinhx)^2????????

4. Feb 13, 2009

### jgens

Re: derivative help!!!!!!!

Still not right. Cosh(x)^2 does not simply cancel out and vanish. If I asked you what 5/5 was equivalent to would you say zero?

To get the correct answer use the identity cosh(x)^2 - sinh(x)^2 = 1 or the identity 1 - tanh(x)^2 = sech(x)^2

5. Feb 13, 2009

### ldbaseball16

Re: derivative help!!!!!!!

ooooooo the identity oo ok so its 1/(coshx)^2?????????????????

6. Feb 13, 2009

### jgens

Re: derivative help!!!!!!!

Can you simplify that further?

7. Feb 13, 2009

### ldbaseball16

Re: derivative help!!!!!!!

no i get zero if i do the quotient rule again.....????

8. Feb 13, 2009

### jgens

Re: derivative help!!!!!!!

No, you don't need to apply the quotient rule again nor should you; moreover, the derivative of 1/cosh(x)^2 is not zero.

Maybe you haven't been taught this but sech(x) = 1/cosh(x). Now simplify it.

9. Feb 13, 2009

### ldbaseball16

Re: derivative help!!!!!!!

oooo yessss i found it in my notebook in the hyperbolic functions would it be sech^2x???????????

10. Feb 13, 2009

### jgens

Re: derivative help!!!!!!!

Yes, d(tanh(x))/dx = sech(x)^2.

11. Feb 14, 2009

### Staff: Mentor

Re: derivative help!!!!!!!

You've come to this Web site, so I suppose that means you would like some help. You've taken the first step, which is to show us what you're tried to do. Good.