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 […]
What is the integration of 1/x log x ? Read More »
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
What is the integration of x log x dx ? Read More »
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
What is the integration of \((log x)^2\) dx ? Read More »
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
What is the differentiation of log sin x ? Read More »
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
What is the differentiation of 1/log x ? Read More »
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
What is the differentiation of \(log x^2\) ? Read More »
Solution : We have, y = x log x By using product rule in differentiation, \(dy\over dx\) = log x \(dy\over dx\)(x) + x \(dy\over dx\) (log x) \(dy\over dx\) = log x + x/x \(\implies\) \(dy\over dx\) = log x + 1 Hence, the differentiation of x log x with respect to x is
What is the Differentiation of x log x ? Read More »
Solution : We have, y = log ( log x ) By using chain rule in differentiation, \(dy\over dx\) = \(1\over log x\).\(1\over x\) \(\implies\) \(dy\over dx\) = \(1\over x log x\) Hence, the differentiation of log log x with respect to x is \(1\over x log x\). Similar Questions What is the differentiation of
What is the Differentiation of log log x ? Read More »
Solution : On solving the equations 4x + y – 1 = 0 and 7x – 3y – 35 = 0 by using point of intersection formula, we get x = 2 and y = -7 So, given lines intersect at (2, -7) Now, the equation of line joining the point (3, 5) and (2,
Solution : Solving simultaneously the equations 2x – y + 3 = 0 and x + y – 5 = 0, we obtain \(x\over {5 – 3}\) = \(y\over {3 + 10}\) = \(1\over {2 + 1}\) \(\implies\) \(x\over 2\) = \(y\over 13\) = \(1\over 3\) \(\implies\) x = \(2\over 3\) , y = \(13\over
