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Continuity And Differentiability

Question
CBSEENMA12034877

If u, v, w are differentiable function ofx, then show that
straight d over dx left parenthesis straight u. straight v. straight w right parenthesis equals du over dx. vw plus straight u. dv over dx. straight w plus straight u. straight v dw over dx
in two ways - first by repeated application of product rule, second by logarithmic differentiation.

Solution
Let space space space space space space space space space space space space space space space space space straight y equals straight u. straight v. straight w
therefore space space space space space space space space space space space space space dy over dx equals straight d over dx left parenthesis straight u. straight v. straight w right parenthesis equals straight d over dx left square bracket left parenthesis uv right parenthesis. straight w right square bracket equals uv. dw over dx plus straight w straight d over dx left parenthesis uv right parenthesis
space space space space space space space space space space space space space space space space space space space space space space space space space equals uv dw over dx plus straight w open parentheses straight u dv over dx plus straight v du over dx close parentheses
space space space space space space space space space space space space space space space space space space space space space space space space space equals uv dw over dx plus wu dv over dx plus vw du over dx
therefore space straight d over dx left parenthesis straight u. straight v. straight w right parenthesis equals du over dx straight v. straight w plus straight u. dv over dx. straight w plus uv. dw over dx space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis
Again space let space straight y equals straight u. straight v. straight w
therefore space space space space space space space space space space space space space log space straight y equals log left parenthesis straight u. straight v. straight w right parenthesis
therefore space space space space space space space space space space space space space log space straight y equals log space straight u plus log space straight v plus log space straight w
therefore space space space space space space space space space 1 over straight y dy over dx equals fraction numerator begin display style du over dx end style over denominator straight u end fraction plus fraction numerator begin display style dv over dx end style over denominator straight v end fraction plus fraction numerator begin display style dw over dx end style over denominator straight w end fraction
therefore space space space space space space space space space space space space space space dy over dx equals straight y open square brackets fraction numerator begin display style du over dx end style over denominator straight u end fraction plus fraction numerator begin display style dv over dx end style over denominator straight v end fraction plus fraction numerator begin display style dw over dx end style over denominator straight w end fraction close square brackets
space space space space space space space space space space space space space space space space space space space space space space space space space equals uvw open square brackets fraction numerator begin display style du over dx end style over denominator straight u end fraction plus fraction numerator begin display style dv over dx end style over denominator straight v end fraction plus fraction numerator begin display style dw over dx end style over denominator straight w end fraction close square brackets
therefore space fraction numerator straight d over denominator d x end fraction left parenthesis straight u. straight v. straight w right parenthesis equals fraction numerator d u over denominator d x end fraction. v w plus straight u. fraction numerator d v over denominator d x end fraction. straight w plus straight u. straight v fraction numerator d w over denominator d x end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis
From (1) and (2), result is clear.

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