Cambridge Additional Mathematics

362 Introduction to differential calculus (Chapter 13)

THE DERIVATIVE OF tanx

Consider y= tanx=

sinx

cosx

We let u= sinx and v= cosx

)

du

dx

= cosx and

dv

dx

=¡sinx

)

dy

dx

=

u^0 v¡uv^0

v^2

fquotient ruleg

=

cosxcosx¡sinx(¡sinx)

[cosx]^2

=

cos^2 x+ sin^2 x

cos^2 x

=

1

cos^2 x

fsince sin^2 x+ cos^2 x=1g

= sec^2 x

Function Derivative

sinx cosx

cosx ¡sinx

tanx sec^2 x

THE DERIVATIVES OF sin[f(x)], cos[f(x)], AND tan[f(x)]

Suppose y= sin[f(x)]

If we let u=f(x), then y= sinu.

But

dy

dx

=

dy

du

du

dx

fchain ruleg

)

dy

dx

= cosu£f^0 (x)

= cos[f(x)]£f^0 (x)

We can perform the same procedure for cos[f(x)] and tan[f(x)], giving the following results:

Function Derivative

sin[f(x)] cos[f(x)]f^0 (x)

cos[f(x)] ¡sin[f(x)]f^0 (x)

tan[f(x)] sec^2 [f(x)]f^0 (x)

Example 17 Self Tutor

Differentiate with respect tox:

a xsinx b 4 tan^2 (3x)

a If y=xsinx

then by the product rule

dy

dx

= (1) sinx+(x) cosx

= sinx+xcosx

b If y= 4 tan^2 (3x)

= 4[tan(3x)]^2

=4u^2 where u= tan(3x)

dy

dx

=

dy

du

du

dx

fchain ruleg

)

dy

dx

=8u£

du

dx

= 8 tan(3x)£3 sec^2 (3x)

= 24 sin(3x) sec^3 (3x)

DERIVATIVE

DEMO

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Additional Mathematics
Y:\HAESE\CAM4037\CamAdd_13\362CamAdd_13.cdr Tuesday, 7 January 2014 9:55:05 AM BRIAN