Find the exact coordinates of the stationary point on the curve y = e 2 x x 2 and determine its nature.

Given curve is y = e 2 x x 2 We need to find the gradient of this curve i.e., d y d x d y d x = d d x e 2 x x 2 = x 2 d d x e 2 x - e 2 x d d x x 2 x 4 ∵ d d x u v = v d u d x - u d v d x v 2 = x 2 × e 2 x × d d x 2 x - e 2 x × 2 x x 4 ∵ d d x e a x = e a x d d x a x , d d x x n = n x n - 1 , n ∈ R = x 2 e 2 x × 2 - 2 x e 2 x x 4 = 2 x e 2 x x - 1 x 4 . . . . . i Now, for the stationary points, we have d y d x = 0 ⇒ x x - 1 = 0 ⇒ x = 1 [Since y = e 2 x x 2 is not defined at x = 0 ] The value of y when x = 1 is y = e 2 1 1 2 = e 2 Hence, the exact coordinates of the stationary point on the given curve are 1 , e 2 . Now, let us find whether this is a point of minima or maxima d 2 y d x 2 = d d x d y d x = d d x 2 e 2 x x 2 - 2 e 2 x x 3 [using i ] = 2 d d x e 2 x x 2 - 2 d d x e 2 x x 3 = 2 x 2 d d x e 2 x - e 2 x d d x x 2 x 4 - 2 x 3 d d x e 2 x - e 2 x d d x x 3 x 6 = 2 x 2 e 2 x × 2 - e 2 x × 2 x x 4 - 2 x 3 × e 2 x × 2 - e 2 x × 3 x 2 x 6 = 4 x e 2 x x - 1 x 4 - 2 x 2 e 2 x 2 x - 3 x 6 = 4 e 2 x x - 1 x 3 - 2 e 2 x 2 x - 3 x 4 . . . . i i Now, at x = 1 let us check the sign of d 2 y d x 2 d 2 y d x 2 = 0 - 2 e 2 - 1 1 = 2 e 2 > 0 [using i i ] Thus, 1 , e 2 is a point of local minima.

By writing 2 as e ln 2 prove that d d x 2 x = 2 x ln 2

To Prove: d d x 2 x = 2 x ln 2 Consider d d x 2 x = d d x e ln 2 x ∵ 2 = e ln 2 = d d x e x ln 2 = e x ln 2 × d d x x ln 2 ∵ d d x e a x = e a x d d x a x = e x ln 2 × ln 2 ∵ d d x x n = n x n - 1 , n ∈ R = e ln 2 x × ln 2 = 2 x ln 2 ∵ 2 = e ln 2 Hence, Proved.