DIFFERENTIATION AND INTEGRATION OF HYPERBOLIC FUNCTIONS

Since: sinh x = 1212 ( e x – e – x ) = cosh x Using the quotient rule we can show also: i.e. y = sinh x dy d x = cosh x Similarly: dy d x = sinh x y = cosh x y = tanh x dy d x = sech 2 x ( sinh x ) = d d x 1212 ( e x + e – x )

Example 1: Find dy d x given a) y = sinh 2 x b) y = sinh 5x c) y = sech x a) y = ( sinh x ) 2 2 ( sinh x ) 1 ( cosh x ) dy d x = = sinh 2x b) y = sinh u, where u = 5x d yd xd yd x = dy d u × d ud xd ud x cosh u × 5 = 5cosh5x dy d x = c) y = cosh x 1 dy d x = cosh x (0) – 1 (sinh x) ( cosh x ) 2 = – tanh x sech x

tanh3x dx Example 2: Find a) 2 0 sinh 2x dx b) 2 0 sinh 2x dx a) = 1 2    cosh 2x    2 0 tanh3x dx b) = 1 2 ( cosh 4 – cosh 0 ) = 1 2 ( cosh 4 – 1 ) = ( e 4 + e – 4 )     – 1 = 1 4 ( e 4 + e – 4 – 2 ) = cosh3x sinh3x dxdx = 1 3 ln(cosh3x) + c Results for integration follow from the differentiation results:

Inverse hyperbolic functions If y = sinh – 1 x, then x = sinh y d xd yd xd y = sinh 2 y + 1 cosh y = d xd yd xd y = x dy d x = dxdx 1 x = sinh – 1 x + c dxdx 1 x 2 – 1 = cosh – 1 x + c Similarly:

We can also show: dxdx = tanh – 1 x + c 1 1 – x 2 The results can be extended to more general results: dxdx 1 x 2 + a 2 = sinh – 1 x a + c dxdx 1 x 2 – a 2 = cosh – 1 x a + c dxdx = tanh – 1 1 a 2 – x 2 1 a x a + c

Example 3: Evaluate dxdx 1 9x dxdx = 1313 dxdx 1 x = 1313   sinh – 1 2 3x3x   = 2 3 Using: dxdx 1 x 2 + a 2 = sinh – 1 x a + c

Summary of key points: This PowerPoint produced by R.Collins ; Updated Aug y = sinh x dy d x = cosh x dy d x = sinh x y = cosh x y = tanh x dy d x = sech 2 x dxdx 1 x 2 + a 2 = sinh – 1 x a + c dxdx 1 x 2 – a 2 = cosh – 1 x a + c dxdx = tanh – 1 1 a 2 – x 2 1 a x a + c