Class 12 Maths Continuity Differentiability Logarithmic Differentiation

Logarithmic Differentiation:

Let we have a function in the form y = f(x) = [u(x)]v(x)

Taking logarithm (to base e) on both sides, we get

log y = log [u(x)]v(x)

log y = v(x) * log [u(x)]

Differentiate w.r.t. x using chain rule, we get

      (1/y) * dy/dx = v(x) * 1/u(x) * u’(x) + v’(x) * log[u(x)]

=> dy/dx = y[{v(x)/u(x)} * u’(x) + v’(x) * log u(x)]

Here, f(x) and u(x) must always be positive otherwise their logarithms are not defined. This process of differentiation is known as logarithms differentiation.

Problem: Differentiate the given function w.r.t. x

cos x * cos 2x * cos 3x

Answer:

Let y = cos x * cos 2x * cos 3x

Taking logarithm on both the sides, we obtain

      log y = log(cos x * cos 2x * cos 3x)

=> log y = log cos x + log cos 2x + log cos 3x

Differentiating both sides with respect to x, we get

=> (1/y) * dy/dx = 1/cos x * d(cos x)/dx + 1/cos 2x * d(cos 2x)/dx + 1/cos 3x * d(cos x3)/dx

=> (1/y) * dy/dx = 1/cos x * (-sin x) + 1/cos 2x * (-sin 2x) * d(2x)/dx + 1/cos 3x * (-sin 3x) *  d(3x)/dx

=> (1/y) * dy/dx = -sin x /cos x – 2 * sin 2x /cos 2x – 3 * sin 3x/cos 3x

=> dy/dx = y[-sin x /cos x – 2 * sin 2x /cos 2x – 3 * sin 3x/cos 3x]

=> dy/dx = -cos x * cos 2x * cos 3x [tan x + 2 tan 2x + 3 tan 3x]

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