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 ? …
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 …
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 …
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 …
Solution : We have, I = \(\int\) \(1\over x log x\) dx Put log x = t \(\implies\) \(1\over x\) dx = dt I = \(\int\) \(1\over t\) dt I = log | t | + C I = log |log x| + C Hence, the integration of \(1\over x log x\) is log (log …
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 …
Solution : We have, I = \((log x)^2\) . 1 dx, Then , where \((log x)^2\) is the first function and 1 is the second function according to ilate rule, I = \((log x)^2\) { \(\int\) 1 dx} – \(\int\) {\(d\over dx\) \((log x)^2\) . \(\int\) 1 dx } dx I = \((log x)^2\) 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 …
Solution : We have, y = \(1\over log x\) By using quotient rule in differentiation, \(dy\over dx\) = \(log x.{d\over dx}(1) – 1 {d\over dx}(log x)\over (log x)^2\) \(dy\over dx\) = \(0 – {1\over x}\over (log x)^2\) = \(-1\over x (log x)^2\) Hence, the differentiation of 1/log x with respect to x is \(-1\over x …
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 …
