Solution : The given parabola is of form \(y^2\) = 4ax. On comparing, we have 4a = 12 i.e a = 3. We know that the focal distance of any point (x, y) on \(y^2\) = 4ax is x + a. Let the given point on the parabola \(y^2\) = 12 x be (x, y). …
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 …
What is the differentiation of sin square x or \(sin^2x\) ? Read More »
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 …
