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)
√(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 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 2x [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)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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