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} โ \(\int\) {\(d\over dx\) \((log x)^2\) . \(\int\) 1 dx } dx I = \((log x)^2\) x โฆ
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 \(\implies\) \(dy\over du\) = \(1\over u\)ย Now, \(dy\over dx\) = \(dy\over du\) \(\times\) \(du\over dx\) \(\implies\) \(dy\over dx\) = \(1\over u\) \(\times\) cos x โฆ
Solution : We have, y = \(1\over log x\) By using quotient rule in differentiation, \(dy\over dx\) = \(log x.{d\over dx}(1) โ 1 {d\over dx}(log x)\over (log x)^2\) \(dy\over dx\) = \(0 โ {1\over x}\over (log x)^2\) = \(-1\over x (log x)^2\) Hence, the differentiation of 1/log x with respect to x is \(-1\over x โฆ
Solution : We have y = \(log x^2\) By using chain rule in differentiation, let u = \(x^2\) \(\implies\)ย \(du\over dx\) = 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 โฆ
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 โฆ
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 โฆ
Solution : On solving the equations 4x + y โ 1 = 0 and 7x โ 3y โ 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, โฆ
Solution : Solving simultaneously the equations 2x โ y + 3 = 0 and x + y โ 5 = 0, we obtain \(x\over {5 โ 3}\) = \(y\over {3 + 10}\) = \(1\over {2 + 1}\) \(\implies\) \(x\over 2\) = \(y\over 13\) = \(1\over 3\) \(\implies\) x = \(2\over 3\) , y = \(13\over โฆ
Solution : On solving the equations x โ 7y + 5 = 0 and 3x + y = 0 by using point of intersection formula, we get x = \(-5\over 22\) and y = \(15\over 22\) So, given lines intersect at \(({-5\over 22}., {15\over 22})\) Let the equation of the required line be x = โฆ
Solution : Inner Radius (r) = \(24\over 2\) = 12 cm Outer Radius (R) = \(28\over 2\) = 14 cm Height of Pipe = 35 cm Volume = \(\pi (R^2 โ r^2) h\) = \(\pi \times 52 \times 35\) = 5720 \(cm^3\) Mass of 1 \(cm^3\) wood = 0.6 kg Mass of 5720 \(cm^3\) wood โฆ
