# derivative of cosh

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• October 9th 2008, 03:58 PM
Cucc1137
derivative of cosh
I have this question and i figured an answer but I am not sure if its completely right... Someone help?

the question is

if f(x) = cosh(ln(x)), then show that f ' (x) = 1/2( (x^2 - 1) / x^2)

The way i went about it was i turned the last part into

• 1/2 ( ((x^2)/x^2) - ( 1/x^2) )

• 1/2(1- (1/x^2)

then i took the derivative of f(x)

f ' (x) = sinh(lnx) (1/x)

= (e ^ lnx - e ^ -lnx) / 2 * 1/x

= (x - x^ - 1) / 2 * 1/x

= 1/2 (1 - x^ -2)

= 1/2 ( 1 - (1 / x^2) )

Thanks for the help...
• October 9th 2008, 04:53 PM
mr fantastic
Quote:

Originally Posted by Cucc1137
I have this question and i figured an answer but I am not sure if its completely right... Someone help?

the question is

if f(x) = cosh(ln(x)), then show that f ' (x) = 1/2( (x^2 - 1) / x^2)

The way i went about it was i turned the last part into

• 1/2 ( ((x^2)/x^2) - ( 1/x^2) )

• 1/2(1- (1/x^2)

then i took the derivative of f(x)

f ' (x) = sinh(lnx) (1/x)

= (e ^ lnx - e ^ -lnx) / 2 * 1/x

= (x - x^ - 1) / 2 * 1/x

= 1/2 (1 - x^ -2)

= 1/2 ( 1 - (1 / x^2) )

Thanks for the help...

Your solution is correct.
• October 9th 2008, 04:54 PM
Nacho
the answer is right, however you can use the definition of cosh before

$
\cosh \left( {\ln x} \right) = \frac{{\exp \left\{ {\ln x} \right\} + \exp \left\{ { - \ln x} \right\}}}
{2} = \frac{{x + \frac{1}
{x}}}
{2}
$

there you don´t use de the quotient rule