$f(x) = {\cosh x \over \sinh x}f'(x)$

Question

show $f(x) = {\cosh x \over \sinh x}f'(x) \Rightarrow f(x) =\cosh x$

I can easily check that LHS is well defined with the equation of $f(x) =\cosh x$ as given.

However, just replace the LHS given RHS value cannot be the proof I think.

Thus I had transformed the LHS as below:

${f(x) \over f'(x)} = [\log(f(x))]' = {\cosh x \over \sinh x} = {e^x - e^x \over e^x + e^x} $

but from the last term of above, how could one get $\int{e^x - e^x \over e^x + e^x}$ to get the $\log(f(x))$? any hint?

It should be $(\dots) ; \Rightarrow f(x) = \color{red}{C} \cosh x$ for any constant $C$. — StackTD, Jun 14 '17 at 10:53

score 2 · Answer 1 · answered Jun 14 '17 at 10:58

2

Let $g(x):=\frac{f(x)}{\cosh x}$.

Show that $g'(x)=0$ for all $x$.

Your turn !

answered Jun 14 '17 at 10:58

Fred

Deepak · Answer 2 · 2017-06-14T11:07:41.607

2

$f(x) = {\cosh x \over \sinh x}f'(x)$

Rearrange,

$\frac{f'(x)}{f(x)} = \frac{\sinh x}{\cosh x}$

Integrate both sides wrt $x$,

$\int \frac{f'(x)}{f(x)} dx=\int\frac{\sinh x}{\cosh x}dx$

$\ln f(x) = \ln\cosh x + \ln C$ (putting constant as $\ln C$ allows for a simple $C$ in the final expression).

Since the real logarithm is one to one,

$f(x) = C\cosh x$

edited Jun 14 '17 at 11:07

answered Jun 14 '17 at 11:01

Deepak

score 2 · Accepted Answer · answered Jun 14 '17 at 11:05

2

$$\int \frac{e^x-e^{-x}}{e^x+e^{-x}}dx=\int \frac{1}{e^x+e^{-x}}d(e^x+e^{-x})=\ln(e^x+e^{-x})+C$$

answered Jun 14 '17 at 11:05

farruhota

you applied the method of substitution almost correctly. see: $\color{red}t = e^x + e^{-x}, d\color{red}t = (e^x - e^{-x})d\color{red}x,$ then the integral is $\int \frac{(e^x-e^{-x})dx}{e^x+e^{-x}}=\int\frac{dt}t.$ – farruhota Jun 14 '17 at 11:30

3 Answers3