D.w.r.t.x (cosh x)^(sin x) = Derivative of (cosh x)^(sin x)

[DryRun]

Homework Statement
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.

 

[Dick]

Homework Statement
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.

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.

 

[DryRun]

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.

 

[DryRun]

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?

 

[Dick]

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.

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

 

[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:
$g(x)f(x)^{g(x)- 1}f'(x)+ f(x)^{g(x)}g'(x)ln(f(x)$

 

[SammyS]

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

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

 

[Dick]

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

Thank you, SammyS.