Gradients of Inverse Trig FunctionsUse the relationship dy 1

dxdxdy

Ex y = sin–1xThis is the same as siny = x

ie sin both sidesSo x = siny

Differentiate this expression. Remember the letter that comes 1st goes on top

The letter that comes 2nd goes underneath.

dxcos y

dy

x = siny

dy 1 1dxdx cos ydy

Using sin2x + cos2x = 1 replace y by x. So sin2y + cos2y = 1

Make cosy the subject cos2y = 1 – sin2y

2cos y 1 sin y

But x = siny So 2cos y 1 x

2

dy 1

dx 1 x

This technique can be used to differentiate y = cos–1x and y = tan–1x (use 1 + tan2x = sec2x)

Find the gradient of y = cos–1x and y = tan–1x

x xe e

2

x xdy e e

dx 2

x xe e

2

x xdy e e

dx 2

2dysech x

dx

y = sinh x =

Differentiating Hyperbolic Functions

y = cosh x =

y = tanh x =sinhx

coshx

x x

x x

e e

e e

Prove it using the quotient rule

coshx

sinhx

dy2cosh2x

dx

Using techniques from C4 (DIFIU)

y = cosh x

y = sinh 2x

dy 1 1

sinh xdx 3 3

y = tanh (3x+2) 2dy3sech 3x 2

dx

dxcoshy

dy

dy 1

dx coshy

Differentiating Inverse Hyperbolic Functions

y = sinh–1x so x = sinhy

cosh2x – sinh2x = 1Using Osbornes Rule

2 2

1 1

1 sinh y 1 x

dx

sinhydy

dy 1

dx sinhy

Differentiating Inverse Hyperbolic Functions

y = cosh–1x so x = coshy

cosh2x – sinh2x = 1Using Osbornes Rule

2 2

1 1

cosh y 1 x 1

2dxsech y

dy

2

dy 1

dx sech y

Differentiating Inverse Hyperbolic Functions

y = tanh–1x so x = tanhy

1 – tanh2x = sech2x Using Osbornes Rule

2 2

1 1

1 tanh y 1 x

See FP2 Notes for further examples