LOGARITHMIC DIFFERENTIATION PROBLEMS AND SOLUTIONS

Differentiate each of the following with respect to x.

Problem 1 :

x^y = y^x 

Solution :

x^y = y^x 

Taking logarithm on both sides.

logx^y = logy^x

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)

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

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

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

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

Problem 5 :

y = √(x² + 4) sin²x 2^x

Solution :

y = √(x² + 4) sin²x 2^x

Taking logarithms on both sides

log y = log [ √(x² + 4) sin²x 2^x]

Decomposing using properties of logarithms.

= log √(x² + 4) + log sin²x + log 2^x

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

= (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 [x/(x² + 4)] + 2 cot x + log 2

Problem 6 :

y = x^3/4 √(x² + 1) / (3x + 2)⁵

Solution :

y = x^3/4 √(x² + 1) / (3x + 2)⁵

Taking logarithms on both sides

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

log y = log [x^3/4] + 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)]]

= x^3/4 √(x² + 1) / (3x + 2)⁵ [(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)

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