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What is the differentiation of 1/sinx ?

Solution : We have, y = 1/sinx = cosecx By using differentiation formula of cosecx, \(dy\over dx\) = -cosecx.cotx Hence, the differentiation of 1/sinx = -cosecx.cotx Questions for Practice What is the differentiation of \(e^{sinx}\) ? What is the differentiation of sin square x or \(sin^2x\) ? What is the differentiation of cosx sinx ? …

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What is the differentiation of \(sin x^2\) ?

Solution : We have, y = \(sin x^2\) Differentiating with respect to x by using chain rule, \(dy\over dx\) = \(cos x^2\).(2x) \(dy\over dx\) = 2x.\(cos x^2\) Hence, the differentiation of \(sin x^2\) with respect to x is 2x.\(cos x^2\) Questions for Practice What is the differentiation of \(e^{sinx}\) ? What is the differentiation of …

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What is the integration of log cos x dx ?

Solution : We have, I = \(\int\) log cos x dx By using integraton by parts, I = \(\int\) 1.log cos x dx Taking log cos x as first function and 1 as second function. Then, I = log cos x \(\int\) 1 dx – \(\int\) { \({d\over dx}\) (log cos x) \(\int\) 1 dx …

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What is the integration of log 1/x ?

Solution :  We have, I = \(\int\) \(log {1\over x}\) dx I = \(\int\) \(log 1 – log x\) dx = \(\int\) (-log x) dx By using integration by parts formula, Let I = -(\(\int\) log x .1) dx where log x is the first function and 1 is the second function according to ilate …

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What is the integration of x log x dx ?

Solution : We have, I = \(\int\) x log x dx By using integration by parts, And taking log x as first function and x as second function. Then, I = log x { \(\int\) x dx } – \(\int\) { \({d\over dx}(log x) \times \int x dx\) } dx I = (log x) \(x^2\over …

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What is the differentiation of log sin x ?

Solution : We have, y = log sin x By using chain rule in differentiation, Let u = sin x \(\implies\) \(du\over dx\) = cos x And, y = log u \(\implies\) \(dy\over du\) = \(1\over u\)  Now, \(dy\over dx\) = \(dy\over du\) \(\times\) \(du\over dx\) \(\implies\) \(dy\over dx\) = \(1\over u\) \(\times\) cos x …

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What is the differentiation of \(log x^2\) ?

Solution : We have y = \(log x^2\) By using chain rule in differentiation, let u = \(x^2\) \(\implies\)  \(du\over dx\) = 2x And, y = log u \(\implies\) \(dy\over du\) = \(1\over u\) = \(1\over x^2\) Now, \(dy\over dx\) = \(dy\over du\) \(\times\) \(du\over dx\) \(\implies\) \(dy\over dx\) = \(1\over u\).\(du\over dx\) \(\implies\) \(dy\over …

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