# Simple Question - Derivative of log(coshx-1)

1. May 3, 2017

### Ryansf98

1. The problem statement, all variables and given/known data
f(x) = Log(cosh(x-1)), find f'(x).

2. Relevant equations

3. The attempt at a solution
f'(x) = [1/cosh(x) - 1] * d/dx [cosh(x) - 1], => f'(x) = sinh(x) / [cosh(x) - 1]

Although, my marking scheme says the answer should instead be; cosh(x) + 1 / sinh(x).

Can someone explain where I'm going wrong? Thanks.

2. May 3, 2017

### ehild

Check the function f(x) in the problem. Is it really $f(x)=\log(\cosh(x-1))$?

3. May 3, 2017

### Ryansf98

Question 16. Definitely yeah.

4. May 3, 2017

### ehild

log(cosh(x)-1) is not the same as log(cosh(x-1)). The argument of cosh is x in the original problem, you wrote(x-1).
You made errors with the parentheses in the solution. Somehow you wrote the correct derivative, but the marking scheme is not correct.

5. May 3, 2017

### Ryansf98

Ah apologies, I tend to overuse brackets & at times make mistakes as a result. I'm working from a University past paper so I assumed the marking scheme to be correct no matter what as it's a paper from 2014.

So in either circumstances with the parenthesis, there is no way that function could have a derivative equal to the marking schemes answer?

6. May 3, 2017

### ehild

The derivative of f(x)=log(cosh(x)-1) is $\frac{sinh(x)}{\cosh(x)-1}$ as you wrote, and that is not among the given answers.

7. May 3, 2017

### Staff: Mentor

Overuse is not the problem here -- [1/cosh(x) - 1] -- in the righthand side of the first equation above.
What you wrote means $\frac 1 {\cosh(x)} - 1$, which I'm sure isn't what you intended.

When you write a fraction where either the numerator or the denominator contains multiple terms, you have to surround that part with parentheses or other enclosing symbols.