LOGARITHMIC DIFFERENTIATION EXAMPLES

Find derivative of each of the following with respect to x.

Example 1 :

y = x^cosx

Solution :

y = x^cosx

Take logarithm on both sides. 

logy = logx^cosx

logy = cosxlogx

Find derivative with respect to x. 

(1/y)(dy/dx) = cosx(1/x) + logx(-sinx)

dy/dx = y[(cosx/x) - sinxlogx]

dy/dx = x^cosx[(cosx/x) - sinxlogx]

Example 2 :

y = x^logx + (log x)^x

Solution :

y = x^logx + (log x)^x

Let a = x^logx and b = (log x)^x

In (1), find derivative with respect to x.

(1/a)(da/dx) = 2logx(1/x)

 da/dx = 2alogx/x

In (1), find derivative with respect to x.

(1/b)(db/dx) = x(1/logx)(1/x) + log(logx)(1)

db/dx = b[(1/logx) + log(logx)]

y = x^logx + (log x)^x

y = a + b

dy/dx = da/dx + db/dx

dy/d = 2alogx/x + b[(1/logx) + log(logx)]

Substitute a = x^logx and b = (logx)^x.

dy/dx = (2x^logxlogx)/x] + (logx)^x[logx + log(logx)]

Example 3 :

√(xy) = e^{x - y}

Solution :

√(xy) = e^{x - y}

√x√y = e^{x - y}

Take logarithm on both sides.

log(√x√y) = loge^{x - y}

log√x + log√y = (x - y)loge

logx^1/2 + logy^1/2 = x - y

(1/2)logx + (1/2)logy = x - y

Find derivative with respect to x.

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

1/2x + (1/2y)(dy/dx) = 1 - dy/dx

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

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

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

dy/dx = (y/x)[(2x - 1)/(1 + 2y)]

Example 4 :

y = 4^{4x^4}

Solution :

y = 4^{4x^4}

Take logarithm on both sides. 

log y = log(4^{4x^4})

log y = 4x⁴ log 4

Find derivative with respect to x. 

(1/y)(dy/dx) = 4x⁴ (0) + log 4 (4x³)

(1/y)(dy/dx) = 4x³ log 4

dy/dx = y (4x³ log 4)

dy/dx = 4^{4x^4}(4x³ log 4)

Example 5 :

y = e^{x cos x}

Solution :

y = e^{x cos x}

Take logarithm on both sides. 

log y = log(e^{x cos x})

log y = x cos x log(e)

log y = x cos x

Differentiating with respect to x, 

dy/dx = x(-sin x) + cos x (1)

= -x sin x + cos x

Example 6 :

y = log (tan x)

Solution :

y = log (tan x)

Differntiating with respect to x, we get

dy/dx = (1/tan x) sec²x

= (cos x/sin x) ⋅ (1/cos²x)

= 1/sin x cos x 

= cosec x sec x

Example 7 :

y = log (cos (log x))

Solution :

y = log (cos (log x))

Differntiating with respect to x, we get

dy/dx = (1/(cos (log x)) sin (log x) (1/x)

= (1/x) sin (log x) / cos (log x)

= (1/x) tan (log x)

Example 8 :

y = log (sec x + tan x)

Solution :

y = log (sec x + tan x)

Differntiating with respect to x, we get

dy/dx = 1/(sec x + tan x) (sec x tan x + sec²x)

= 1/(sec x + tan x) (sec x (tan x + sec x))

= (sec x (tan x + sec x))/(sec x + tan x) 

= sec x

Example 9 :

y = log (sin (log x))

Solution :

y = log (sin (log x))

Differntiating with respect to x, we get

dy/dx = 1/(sin (log x)) [cos (log x) (1/x)]

= (1/x) [cos (log x)/(sin (log x)]

= (1/x) cot x

Example 10 :

If x^y = e^{x - y} prove that dy/dx = log x / (1 + log x)²

Solution :

If x^y = e^{x - y}

Taking logarithms on both sides

log x^y = e^{x - y}

y log x = (x - y) log e

y log x = (x - y)

y log x + y = x

y(1 + log x) = x

y = x/(1 + log x)

Differentiating with respect to x,

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

 log x(dy/dx) + (dy/dx) = 1 - (y/x)

Applying the value of y, we get

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

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

= (1 + log x - 1)/(1 + log x)

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

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

= log x / (1 + log x)²

Hence proved.

Example 11 :

y = (tan x)^{cot x} 

Solution :

y = (tan x)^{cot x} 

Differentiating with respect to x,

y = (tan x)^{cot x}

log y = log [(tan x)^{cot x}]

log y = cot x log (tan x)

(1/y) (dy/dx) = cot x ((1/tanx) sec²x) + log (tan x)
(-csc²x)

= cot x cot x sec²x - csc²x log (tan x)

= cot²x  secx² - csc²log (tan x)

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