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Logarithmic Differentiation

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we will learn to differentiate certain special class of functions given in the form 
y = f(x) = `[u(x)]^(v (x))`
 By taking logarithm (to base e) the above may be rewritten as
log y = v(x) log [u(x)]

Using chain rule we may differentiate this to get
`1/y . (dy)/(dx) = v(x) . 1/(u(x)) . u'(x) +v'(x) . log [u(x)]`

which implies that
`(dy)/(dx) = y [(v(x))/(u(x)) . u'(x) + v'(x) . log[u(x)]]`

The main point to be noted in this method is that f(x) and u(x) must always be positive as otherwise their logarithms are not defined. This process of differentiation is known as logarithms differentiation. 

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Series 3 | Continuity and Differentiability part 28 (Logarithimic Differentiation)

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Continuity and Differentiability part 28 (Logarithimic Differentiation) [00:09:47]
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