Logarithmic Differentiation

In this method logarithmic differentiation we are going to see some examples problems to understand where we have to apply this method.

Example 1:

Differentiate [sin x cos (x²)]/[ x³ + log x ] with respect to x


We can differentiate this function using quotient rule, logarithmic-function. When we apply the quotient rule we have to use the product rule in differentiating the numerator. But in the method of logarithmic-differentiation first we have to apply the formulas log(m/n) = log m - log n and log (m n) = log m + log n.

Let y = [sin x cos (x²)]/[ x³ + log x ]

Take log on both sides

log y = log [sin x cos (x²)]/[ x³ + log x ]

log y = log [sin x cos (x²)] - log [ x³ + log x ]

log y = log [sin x ] + log [ cos (x²) ] - log [ x³ + log x ]

(1/y)dy/dx=(1/sin x)cos x-[1/cos (x²)]Sin (x²)(2x)-[1/(x³ + log x )]( 3x² + 1/x )

(1/y)dy/dx=(cos x/sin x)-(2x)sin (x²)/cos (x²)-[1/(x³+log x )]( 3x³ + 1)/x

dy/dx = [cot x  - (2x) tan (x²)  - ( 3x³ + 1)/x(x³ + log x )]y

dy/dx = [cot x  - (2x) tan (x²)  - ( 3x³ + 1)/x(x³ + log x )] x

           [sin x cos (x²)]/[ x³ + log x ]    logarithmic differentiation

Example 2:

Differentiate [(x² + 2) (x + √2)]/ [(√(x+4) - (x-7)] with respect to x


Let y = [(x² + 2) (x + √2)]/ [(√(x+4) - (x-7)]

log y = log [(x² + 2) (x + √2)]/ [(√(x+4) - (x-7)]

log y = log [(x² + 2) (x + √2)] - log [(√(x+4) - (x-7)]

log y = log (x² + 2) + log  (x + √2) - log [(√(x+4) - (x-7)]

(1/y)dy/dx=[1/(x²+2)]2x+1/(x+√2)-[1/ (√(x+4)-(x-7)]x 1/2√(x+4) - 1

(1/y)dy/dx=[2x/(x²+2)]+1/(x+√2)-{[[1/2√(x+4)] - 1]/ (√(x+4)-(x-7)}

(1/y)dy/dx=[2x/(x²+2)]+1/(x+√2)-{[[1 - 2√(x+4)/2√(x+4)]/ (√(x+4)-(x-7)}

dy/dx=yx [2x/(x²+2)]+1/(x+√2)-{[[1 - 2√(x+4)/2√(x+4)]/ (√(x+4)-(x-7)}

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Logarithmic Differentiation to First Principles

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