Derivatives of Hyperbolic Trigonometric Functions

Suggested Prerequesites:

Intro to Hyperbolic trig Functions, Useful Hyperbolic Trig Identities, Derivatives of exponentials, The Chain rule

Hyperbolic trig functions, although many people discredit them, can actually be very useful. True, there are few examples of explicit hyperbolic functions in the physical world. However, using hyperbolic functions can make exponentials appear to behave like trigonometric functions -- an analogy that can provide much intuition.

In any case, we still want to know the derivatives of the hyperbolic functions. We'll find the derivatives of sinh and cosh from their definitions in terms of exponentials:

d dx sinh (x) &sp;=&sp; d dx (12) (e^x^  -  e^-x^) &sp;=&sp; (12) (e^x^
 +  e^-x^) &sp;=&sp; cosh (x)

$d dx cosh (x) &sp;=&sp; d dx (12) (e^x^ + e^-x^) &sp;=&sp; d dx (12) (e^x^ - e^-x^) &sp;=&sp; sinh (x)$

OK, let's take a look at this. The derivatives of sinh and cosh seem to behave just like the derivatives of sin and cos, except that the annoying negative signs have gone away. (Remember: D_xsin(x) = cos(x), D_xcos(x) = -sin(x)) We can find the derivatives of the other hyperbolic trig functions using the quotient rule, the chain rule and the derivatives we've just derived:

d dx tanh (x) &sp;=&sp; d dx sinh (x) cosh (x) &sp;=&sp; cosh^2^(x)  - sinh^2^(x) cosh^2^(x) &sp;=&sp; 1 cosh^2^(x) &sp;=&sp; sech^2^(x)

$d dx coth (x) &sp;=&sp; d dx cosh (x) sinh (x) &sp;=&sp; cosh^2^(x) - sinh^2^(x) sinh^2^(x) &sp;=&sp; -1 sinh^2^(x) &sp;=&sp; - csch^2^(x)$

$d dx sech (x) &sp;=&sp; d dx (cosh x)^-1^ &sp;=&sp;
(-1)(cosh x)^-2^(sinh x) &sp;=&sp;
- sech (x) tanh (x)$

$d dx csch (x) &sp;=&sp; d dx (sinh x)^-1^ &sp;=&sp;
(-1)(sinh x)^-2^(cosh x) &sp;=&sp;
- csc (x) coth (x)$

So, as with sinh and cosh, the derivatives of the other hyperbolic trig functions closely resemble those of the normal trig functions, with some discrepancies over negative signs. Be careful, though -- those negative signs can easily cause big errors!

Some examples:

$D_x_(sinh^2^x - cosh^2^x)
&sp;=&sp; 2 sinh x&sp; cosh x -
2 cosh x&sp; sinh x &sp;=&sp; 0$
This is the result we would expect since $sinh^2^x
- cosh^2^ &sp;=&sp; 1$
$D_x_(ln (sinh x) ) &sp;=&sp; cosh x sinh x &sp;=&sp; tanh x$

Exercises:

$y &sp;=&sp; e$ ^sinh²x
$y &sp;=&sp; tanh x x$
$y &sp;=&sp; cosh (e^ln x^)$

Solutions to the exercises | Back to the Calculus page | Back to the World Web Math top page

jjnichol@mit.edu