Differentiation Questions

Solution : Let y = \(x^{sinx}\). Then, Taking log both sides, log y = sin x.log x \(\implies\) y = \(e^{sin x.log x}\) By using logarithmic differentiation, On differentiating both sides with respect to x, we get \(dy\over dx\) = \(e^{sin x.log x}\)\(d\over dx\)(sin x.log x) \(\implies\) \(dy\over dx\) = \(x^{sin x}{log x {d\over dx}(sin […]

Solution : Since by deleting a single term from an infinite series, it remains same. Therefore, the given function may be written as y = \(\sqrt{sin x + y}\) Squaring on both sides, \(\implies\)  \(y^2\)  = sin x + y By using differentiation of infinite series, Differentiating both sides with respect to x, 2y \(dy\over

Solution : We have, x = a{cos t + \({1\over 2} log tan^2 {t\over 2}\)} and y = a sin t \(\implies\) x = a{cos t + \({1\over 2} \times 2 log tan{t\over 2}\)} and y = a sin t \(\implies\) x = a{cos t + {\(log tan{t\over 2}\)} and y = a sin t

Solution : Let y = cosx.sinx By using product rule in differentiation, \(dy\over dx\) = sinx(-sinx) + cosx.cosx \(dy\over dx\) = \(cos^2x – sin^2x\) = cos 2x Hence, the differentiation of cosx.sinx with respect to x is cos 2x. Questions for Practice What is the differentiation of \(e^{sinx}\) ? What is the differentiation of sin

Solution : Let y = \(e^{sinx}\). Putting u = sinx , we get y = \(e^u\) and u = sinx \(\therefore\)  \(dy\over du\) = \(e^u\) and \(du\over dx\) = cosx Now, \(dy\over dx\) = \(dy\over du\) \(\times\) \(du\over dx\) \(\implies\) \(dy\over dx\) = \(e^u\)cosx = \(e^{sinx}\)cosx Hence, the differentiation of \(e^{sinx}\) with respect to x

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

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, 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