# D.w.r.t.x (cosh x)^(sin x)

1. Dec 6, 2011

### sharks

The problem statement, all variables and given/known data
D.w.r.t.x (cosh x)^(sin x)

The attempt at a solution
My attempt:

To me, it appears like (cosh x)^(sin x) is the same as a^x, which on differentiating gives me:
a^x. ln a

My answer: (cosh x)^(sin x).(lncosh x) and then i think that's the end of it.
But the true answer is something else.

2. Dec 6, 2011

### Dick

Your formula for the derivative of a^x assumes a is a constant. Rewrite cosh(x)=e^(log(cosh(x)), use the rules of exponents and try that again.

3. Dec 6, 2011

### sharks

So, d.w.r.t.x (cosh x)^(sin x) gives (according to my understanding)

The equivalent expression: e^[ln(cosh x)^(sin x)]

I can relate to e^ax which gives e^ax.(a)

So, in this problem, e^[ln(cosh x)^(sin x)] = e^[ln(cosh x)^(sin x)].(1/(cosh x)^(sin x))
And i'm gloriously stuck again.

4. Dec 6, 2011

### sharks

For some reason, since the past 24 hours, i cannot edit any of my posts anymore as the forum permissions have been changed, so i'll just have to keep adding replies instead of editing my posts...

The part that needs to be completed (i think) is the d.w.r.t.x of (cosh x)^(sin x). But i don't know how to do that. Maybe substitution? Should i substitute cosh x or sin x?

5. Dec 6, 2011

### Dick

(e^(ln(cosh(x)))^sin(x)=e^(ln(cosh(x)*sin(x)). (e^a)^b=e^(a*b), right?

6. Dec 6, 2011

### HallsofIvy

Another example of what I have always considered an amusing property: In differentiating $f(x)^{g(x)}$, there are two mistakes one can make:

1) Treat g(x) as if it were a constant and use the power rule: $g(x)f(x)^{g(x)- 1}f'(x)$

2) Treat f(x) as if it were a constant an do a "logarithmic" derivative: $f(x)^{g(x)}g'(x)ln(f(x)$

The amusing part is that correct derivative is the sum of those two mistakes:
$g(x)f(x)^{g(x)- 1}f'(x)+ f(x)^{g(x)}g'(x)ln(f(x)$

7. Dec 6, 2011

### SammyS

Staff Emeritus
There's a missing parenthesis which I inserted above in red.

Latex may make the above clearer.

$\displaystyle \left(e^{\ln(\cosh(x))}\right)^{\sin(x)}=e^{\ln( \cosh(x))\sin(x)}$

Remember: $\displaystyle \left(e^a\right)^b=e^{ab}$

8. Dec 6, 2011

### Dick

Thank you, SammyS.