Higher Engineering Mathematics

Ess-For-H8152.tex 19/7/2006 18: 2 Page 710

710 ESSENTIAL FORMULAE

yorf(x)

dy

dx

orf′(x)

cosec−^1 f(x)

−f′(x)

f(x)

√

[f(x)]^2 − 1

cot−^1

x

a

−a

a^2 +x^2

cot−^1 f(x)

−f′(x)

1 +[f(x)]^2

sinh−^1

x

a

1

√

x^2 +a^2

sinh−^1 f(x)

f′(x)

√

[f(x)]^2 + 1

cosh−^1

x

a

1

√

x^2 −a^2

cosh−^1 f(x)

f′(x)

√

[f(x)]^2 − 1

tanh−^1

x

a

a

a^2 −x^2

tanh−^1 f(x)

f′(x)

1 −[f(x)]^2

sech−^1

x

a

−a

x

√

a^2 −x^2

sech−^1 f(x)

−f′(x)

f(x)

√

1 −[f(x)]^2

cosech−^1

x

a

−a

x

√

x^2 +a^2

cosech−^1 f(x)

−f′(x)

f(x)

√

[f(x)]^2 + 1

coth−^1

x

a

a

a^2 −x^2

coth−^1 f(x)

f′(x)

1 −[f(x)]^2

Product rule:

Wheny=uvanduandvare functions ofxthen:

dy

dx

=u

dv

dx

+v

du

dx

Quotient rule:

Wheny=

u

v

anduandvare functions ofxthen:

dy

dx

=

v

du

dx

−u

dv

dx

v^2

Function of a function:

Ifuis a function ofxthen:

dy

dx

=

dy

du

×

du

dx

Parametric differentiation

Ifxandyare both functions ofθ, then:

dy

dx

=

dy

dθ

dx

dθ

and

d^2 y

dx^2

=

d

dθ

(

dy

dx

)

dx

dθ

Implicit function:

d

dx

[f(y)]=

d

dy

[f(y)]×

dy

dx

Maximum and minimum values:

Ify=f(x) then

dy

dx

= 0 for stationary points.

Let a solution of

dy

dx

=0bex=a; if the value of

d^2 y

dx^2

whenx=ais:positive, the point is aminimum,

negative, the point is amaximum.

Velocity and acceleration

If distancex=f(t), then

velocity v=f′(t)or

dx

dt

and

acceleration a=f′′(t)or

d^2 x

dt^2