EXAMPLES OF LOGARITHMIC DIFFERENTIATION

Differentiate each of the following with respect to x.

Example 1 :

y = [sinxcos(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.

lny = ln(x√x)

lny = √xln(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) + lnx/2√x

= (1/x) + lnx/2√x

(y'/y) = (2 + lnx)/2√x

y' = x√x(2 + lnx)/2√x

Example 4 :

y = xsinx

Solution :

y = xsinx

Take logarithm on both sides.

lny = lnxsinx

lny = sinxlnx

Differentiate with respect to x.

u = sinx and u' = cosx

v = lnx and v' = 1/x

(1/y)y' = sinx(1/x) + lnx(cosx)

y'/y  =  sinx/x + cosx(lnx)

y' = y[(sinx/x) + cosx(lnx)]

y' = xsinx[(sinx/x) + cosx(lnx)]

Example 5 :

y = (lnx)cosx

Solution :

y = (lnx)cosx

Take logarithm on both sides.

lny = ln((lnx)cosx)

lny = cosxln(lnx)

Differentiate with respect to x.

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

y' = y[cosx/xlnx) - sinx(ln (lnx)]

y' = (lnx)cosx(cosx/xlnx) - sinx(ln(lnx))

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