LOGARITHMIC DIFFERENTIATION PROBLEMS AND SOLUTIONS

Logarithmic differentiation is a method of finding derivatives of some complicated functions, using the properties of logarithms.

There are cases in which differentiating the logarithm of a given function is easier than differentiating the function as it is.

Step 1 : 

Take logarithm on both sides of the given equation. 

Step 2 : 

Use the properties of logarithm. 

Step 3 : 

Differentiate with respect to x and solve for dy/dx. 


Differentiate each of the following with respect to x.

Problem 1 :

xy = yx 

Solution :

xy = yx 

Taking logarithm on both sides.

logxy = logyx

ylogx = xlogy

Differentiate with respect to x. 

y(1/x) + logx(dy/dx) = x(1/y)(dy/dx) + logy(1)

y/x + logx(dy/dx) = (x/y)(dy/dx) + logy

logx(dy/dx) - (x/y)(dy/dx) = logy - y/x

(logx - x/y)(dy/dx) = (xlogy - y)/x

[(ylogx - x)/y](dy/dx) = (xlogy - y)/x

dy/dx = (y/x)[(xlogy - y)/(ylogx - x)]

Problem 2 :

y = (cosx)logx

Solution :

y = (cosx)logx

logy = log[(cosx)logx]

logy = (logx)log(cosx)

Differentiate with respect to x. 

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

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

(dy/dx) = y[-logxtanx + log(cosx)/x]

dy/dx = (cosx)logx[-logxtanx + log(cos x)/x]

Problem 3 :

(x2/a2) + (y2/b2) = 1

Solution :

Its an implicit function. Since the function is not a complicated one, we don't have to use logarithm to find derivative.

(x2/a2) + (y2/b2) = 1

Differentiate with respect to x. 

(2x/a2) + (2y/b2)(dy/dx) = 0

(2y/b2)(dy/dx) = -(2x/a2)

(dy/dx) = -(b2/a2)(x/y)

(dy/dx) = -(b2x/a2y)

Problem 4 :

√(x2 + y2) = tan-1(y/x)

Solution :

√(x2 + y2) = tan-1(y/x)

Differentiate √(x2 + y2) with respect to x.

= [1/2√(x2+ y2)][2x + 2y(dy/dx)]

= [x + y(dy/dx)]/√(x2+y2) ----(1)

Differentiate tan-1(y/x) with respect to x.

  = 1/[1 + (y/x)2](-y/x2) + (1/x)(dy/dx)

= [x2/(x2 + y2)][x(dy/dx) - y]/x2

(x(dy/dx) - y)/(x2 + y2) ----(2)

(1) = (2)

[x +  y(dy/dx)]/√(x2+y2) = (x(dy/dx) - y)/(x2 + y2)

√(x+ y2)[x + y(dy/dx)] = x(dy/dx) - y

x√(x+ y2) + y√(x+ y2)(dy/dx) - x(dy/dx)  =  -y

(dy/dx)(y√(x+ y2) - x) = -y - x√(x+ y2)

(dy/dx) = (x√(x2+y2) + y)/(x - y√(x2+y2))

Problem 5 :

y = √(x2 + 4) sin2x 2x

Solution :

y = √(x2 + 4) sin2x 2x

Taking logarithms on both sides

log y = log [ √(x2 + 4) sin2x 2x]

Decomposing using properties of logarithms.

= log √(x2 + 4) + log sin2x + log 2x

= log (x2 + 4)1/2 + log (sin x)2 + log 2x

= (1/2)log (x2 + 4) + 2log (sin x) + x log 2

Differentiating with respect to x, we get

= (1/2)[1/(x2 + 4)](2x) + 2 (1/sin x) cos x + x(0) + log 2(1)

= [x/(x2 + 4)] + 2 (cos x/sin x) + log 2

(1/y) (dy/dx) = [x/(x2 + 4)] + 2 cot x + log 2

dy/dx = y [x/(x2 + 4)] + 2 cot x + log 2

Applying the value of y here, we get

dy/dx = √(x2 + 4) sin2x 2[x/(x2 + 4)] + 2 cot x + log 2

Problem 6 :

y = x3/4 √(x2 + 1) / (3x + 2)5

Solution :

y = x3/4 √(x2 + 1) / (3x + 2)5

Taking logarithms on both sides

log y = log [x3/4 √(x2 + 1) / (3x + 2)5]

log y = log [x3/4] + log √(x2 + 1) - log (3x + 2)5

log y = (3/4) log x + log (x2 + 1)1/2 - 5 log (3x + 2)

log y = (3/4) log x + (1/2) log (x2 + 1) - 5 log (3x + 2)

Differentiating with respect to x, we get

(1/y)(dy/dx) = (3/4) (1/x) + (1/2) (1/(x2 + 1))(2x) - 5[1/(3x+2)] (3)

= (3/4) (1/x) + (x/(x2 + 1)) - [15/(3x+2)]

dy/dx = y [(3/4x) + (x/(x2 + 1)) - [15/(3x+2)]]

= x3/4 √(x2 + 1) / (3x + 2)[(3/4x) + (x/(x2 + 1)) - [15/(3x+2)]]

Problem 6 :

y = tan-1 [(1 + x) / (1 - x)], find y'

Solution :

Instead of differentiating this directly, we may use the method of subsitution.

Let t = (1 + x) / (1 - x)

Let x = tan θ

t = (1 + tan θ) / (1 - tan θ)

Here 1 = tan π/4

t = (tan π/4 + tan θ) / (1 - tan π/4 tan θ)

= tan (π/4 + θ)

Applying the value of t, we get

y = tan-1 [tan (π/4 + θ)]

y = (π/4 + θ)

Since x = tan θ, θ = tan-1(x)

y = π/4 +  tan-1(x)

Differentiating with respect to x, we get

dy/dx = 0 + (1/1 + x2)

dy/dx = 1/(1 + x2)

Problem 7 :

Fidn f'(x) , if f(x) = cos-1 (4x3 - 3x)

Solution :

f(x) = cos-1 (4x3 - 3x)

Instead of differentiating this directly, we may use the method of subsitution.

Let t = 4x3 - 3x

Let x = cos θ

t = 4(cos θ)3 - 3(cos θ)

4 cos3θ - 3cos θ

t = cos 3θ

Applying the value of t, we get

f(x) = cos-1 (cos 3θ)

= 3θ

x = cos θ

θ = cos-1(x)

f(x) = cos-1 (cos 3θ)

= 3θ

Applying the value of θ

f(x) = 3cos-1(x)

f'(x) = 3(-1/(1 + x2))

f'(x) = -3/(1 + x2)

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.

©All rights reserved. onlinemath4all.com

Recent Articles

  1. Digital SAT Math Problems and Solutions (Part - 167)

    May 22, 25 09:59 AM

    digitalsatmath211.png
    Digital SAT Math Problems and Solutions (Part - 167)

    Read More

  2. AP Calculus AB Problems with Solutions (Part - 23)

    May 21, 25 01:19 PM

    apcalculusab22.png
    AP Calculus AB Problems with Solutions (Part - 23)

    Read More

  3. Digital SAT Math Problems and Solutions (Part - 166)

    May 21, 25 05:33 AM

    digitalsatmath210.png
    Digital SAT Math Problems and Solutions (Part - 166)

    Read More