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
Solution : On solving the equations x – 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 =
Solution : Inner Radius (r) = \(24\over 2\) = 12 cm Outer Radius (R) = \(28\over 2\) = 14 cm Height of Pipe = 35 cm Volume = \(\pi (R^2 – r^2) h\) = \(\pi \times 52 \times 35\) = 5720 \(cm^3\) Mass of 1 \(cm^3\) wood = 0.6 kg Mass of 5720 \(cm^3\) wood
Solution : Lateral or Curved Surface Area of Cylinder = \(2 \pi rh\) \(\implies\) \(2 \pi rh\) = 94.2 \(\implies\) \(2 \pi r \times 5\) = 94.2 \(\implies\) r = 3 cm Given, height = 5 cm Volume of Cylinder = \(\pi r^2 h\) = \(\pi \times 9 \times 5\) = \(3.14 \times 9 \times
Solution : Circumference of the base cylindrical vessel = \(2\pi r\) \(\implies\) \(2\pi r\) = 132 \(\implies\) r = 21 cm Given, height = 25 cm Volume of cylinder = \(\pi r^2 h\) = \(\pi \times {21}^2 \times 25\) = 34650 \(cm^3\) Since 1litre = 1000 \(cm^3\) Therefore, It can hold 34.65 litres of
Solution : We have, I = \(sec^{-1}\sqrt{x}\) dx Let x = \(sec^2t\) dx = \(2sec^2 t tan t\) dt I = t.\(2sec^2 t tan t\) dt u = t and v = \(tan^2 t\) I = \(\int\) u.dv = u.v – \(\int\) v.du = \(t.tan^2 t\) – \(\int\) \(tan^2 t\) dt I = \(t.tan^2 t\)
What is the integration of sec inverse root x ? Read More »
Solution : We have, I = \(cos^{-1}\sqrt{x}\) . 1 dx By Applying integration by parts, Taking \(cos^{-1}\sqrt{x}\) as first function and 1 as second function. Then I = \(cos^{-1}\sqrt{x}\) \(\int\) 1 dx – \(\int\) {\(d\over dx\)\(cos^{-1}\sqrt{x}\) \(\int\) 1 dx } dx I = x\(cos^{-1}\sqrt{x}\) – \(\int\) \(-1\over 2\sqrt{(1-x)}\sqrt{x}\) . x dx I = x\(cos^{-1}\sqrt{x}\) –
What is the integration of cos inverse root x ? Read More »
