EXAMPLES OF LOGARITHMIC DIFFERENTIATION

Differentiate each of the following with respect to x.

Example 1 :

y = [sin x cos (x2)]/[x3 + lnx]

Solution :

Take  logarithm on both sides

lny = ln[sinxcos(x2)]/[x3 + lnx]

lny = ln[sinxcos(x2)] - ln(x+ lnx)

lny = ln(sinx) + lncos(x2) - ln(x+ lnx) ----(1)

Derivative of ln y :

= (1/y)y'

= y'/y

Derivative of ln(sin x) :

= (1/sinx)cosx

= cosx/sinx

= cot x

Derivative of lncos (x2) :

= (1/cosx2) - sinx2(2x)

= -2xtan x2

Derivative of ln(x+ ln x) :

= [1/(x+ lnx)][3x+ 1/x]

= [1/(x+ lnx)][(3x+ 1)/x]

= [(3x+ 1)/x(x+ lnx)]

(1) :

y'/y = cotx - 2xtanx[(3x+ 1)/x(x+ lnx)]

ny' = y[cotx - 2xtanx- [(3x+ 1)/x(x+ lnx)]]

ny' = ([sinxcos(x2)]/[x+ lnx])[cotx - 2xtanx2- [(3x+ 1)/x(x+ lnx)]]

Example 2 :

y = (x+ 2)(x + √2)/√(x + 4)

Solution :

lny = ln{[(x+ 2)(x + √2)]/√(x + 4)

lny = ln(x+ 2) + ln(x + √2) - ln √(x + 4) ----(1)

Derivative of lny :

= (1/y)y'

= y'/y

Derivative of ln(x+ 2) :

= (1/(x+ 2))(2x)

= 2x/(x2+2)

Derivative of ln(x + √2) :

= 1/(x + √2)

Derivative of ln√(x + 4) :

= (1/√(x + 4)) ⋅ 1/2√(x + 4)

=  1/2(x + 4)

(1) :

y'/y = [2x/(x+ 2)] + 1/(x + √2) + [1/2(x + 4)]

y' =  y[2x/(x+ 2)] + 1/(x + √2) + [1/2(x + 4)]

y' = ((x+ 2)(x + √2)/√(x + 4))[2x/(x+ 2)] + 1/(x + √2) +[1/2(x + 4)]

Example 3 :

y = x√x

Solution :

y = x√x

Take logarithm on both sides.

ln y = ln (x√x)

ln y = √x ln (x)

Differentiate with respect to x.

u = √x and u' = 1/2√x 

v = lnx and v' = 1/x

(1/y) y' = √x (1/x) + ln (x)(1/2√x)

= (√x/x) + ln x/2√x

= (1/√x) + ln x/2√x

(y'/y) = (2 + ln x)/2√x

y' = x√x(2 + ln x)/2√x

Example 4 :

y = xsinx

Solution :

y = xsinx

Take logarithm on both sides.

ln y = ln xsinx

ln y = sin x ln x

Differentiate with respect to x.

u = sin x and u' = cos x

v = ln x and v' = 1/x

(1/y)y' = sin x (1/x) + ln x(cos x)

y'/y  =  sin x/x + cos x (ln x)

y' = y[(sin x/x) + cos x(ln x)]

y' = xsinx[(sin x/x) + cos x(ln x)]

Example 5 :

y = (ln x)cosx

Solution :

y = (ln x)cosx

Take logarithm on both sides.

ln y = ln ((ln x)cosx)

ln y = cosx ln (ln x)

Differentiate with respect to x.

(1/y)y' = cos x (1/ln x)(1/x) + ln(ln x)(-sinx)

y' = y[cos x/x ln x) - sin x(ln (ln x)]

y' = (lnx)cosx(cos x/x ln x) - sin x(ln(ln x))

Example 6 :

y = xlog x + (log x)x

Solution :

y = xlog x + (log x)x

Part 1 :

Let y = xlog x

log y = log (xlog x)

log y = log x (log x)

log y = (log x)2

log y = 2(log x)

Differentiating with respect to x, we get

(1/y) dy/dx = 2(1/x)

dy/dx = y(2/x)

Applying the value of y, we get

xlog x(2/x)

Part 2 :

y = (log x)x

Taking log on both sides

log y = log ((log x)x)

log y = x log (log x)

(1/y)(dy/dx) = x (1/log x)(1/x) + log (log x) (1)

dy/dx = y[1/log x + log (log x)]

Applying the value of y, we get

= (log x)[1/log x + log (log x)]

Example 7 :

y = tan-1 (6x/(1 - 9x2))

Solution :

y = tan-1 (6x/(1 - 9x2))

y = tan-1 (2(3x)/(1 - (3x)2))

Let 3x = tan θ

tan 2 θ = 2 tan θ / (1 - tan2θ)

y = tan-1 (2 tan θ/(1 - tan2θ))

y = tan-1 (tan 2 θ)

y = 2 θ

deriving value of θ from 3x = tan θ

θ = tan-1(3x)

y = 2 tan-1(3x)

dy/dx = 2 (1/(1 + (3x)2) (3)

dy/dx = 6/(1 + (3x)2

Example 8 :

y = tan-1 (√(1 - cos x)/(1 + cos x))

Solution :

y = tan-1 (√(1 - cos x)/(1 + cos x))

From cos 2 x = 1 - 2 sin2 x

2sin2 x = 1 - cos 2x

Dividing θ by 2, we get

2sin2 (x/2) = 1 - cos x ----(1)

From cos 2 x = 2 cos2 x - 1

2 cos2 x/2 = 1 + cos x ----(2)

(1) / (2)

(1 - cos x)/(1 + cos x) = 2sin2 (x/2) / 2 cos2 x/2

(1 - cos x)/(1 + cos x) = tan2 (x/2)

Taking square roots on both sides

(1 - cos x)/(1 + cos x) = tan2 (x/2)

(1 - cos x)/(1 + cos x) = tan (x/2)

Applying the above value, we get

y = tan-1 (√(1 - cos x)/(1 + cos x))

y = tan-1 (tan (x/2))

y = x/2

dy/dx = 1/2

Example 9 :

y = sin-1 (3x - 4x3)

Solution :

y = sin-1 (3x - 4x3)

Let x = sin θ

y = sin-1 (3 sin θ - 4(sin3θ))

y = sin-1 (sin 3θ)

y =

Deriving value of θ

θ = sin-1(x)

y = 3sin-1(x)

Differentiating with respect to x, we get

dy/dx = 3(1 / √(1 - x2))

dy/dx = 3/√(1 - x2)

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