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Logarithmic Differentiation

Find dy / dx ...

Question

Find $\frac{d y}{d x}$ of the functions given in question.

$(c o s x)^{y} = c o s y)^{x}$ .

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Solution

Given, $(c o s x)^{y} = (c o s y)^{x}$ ., taking log on both sides, we get

${(c o s x)^{y}} = l o g {(c o s y)^{x}} o r y l o g (c o s x) = x l o g (c o s y) D i f f e r e n t i a t i n g b o t h s i d e s w . r . t . x, w e g e t y \frac{d}{d x} (l o g c o s x) + l o g c o s x \frac{d}{d x} y = x \frac{d}{d x} (l o g c o s y) + l o g c o s y \frac{d}{d x} x (U s i n g p r o d u c t r u l e) y (\frac{1}{c o s x}) (- s i n x) + l o g (c o s x) \frac{d y}{d x} = x (\frac{1}{c o s y}) (- s i n y) \frac{d y}{d x} + l o g (c o s y) .1 \Rightarrow l o g (c o s s) \frac{d y}{d x} + x t a n y \frac{d y}{d x} = l o g (c o s y) + y t a n x \Rightarrow (l o g (c o s x) + x t a n y) \frac{d y}{d x} = l o g (c o s y) + y t a n x \Rightarrow \frac{d y}{d x} = \frac{l o g (c o s y) + y (t a n x)}{l o g (c o s x) + x t a n y}$

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