Differentiate: a x - b x x

We have to differentiate a x - b x x . So, d d x a x - b x x = x d d x a x - b x - a x - b x d d x x x 2 [Using Quotient rule of differentiation] = x a x log a - b x log b - a x - b x 1 x 2 ∴ d d x a x - b x x = x a x log a - b x log b - a x - b x x 2 .

Differentiate: e x + e - x x 2 + 1

We have to differentiate e x + e - x x 2 + 1 . So, d d x e x + e - x x 2 + 1 = x 2 + 1 d d x e x + e - x - e x + e - x d d x x 2 + 1 x 2 + 1 2 [Using Quotient rule of differentiation] = x 2 + 1 e x + e - x d d x - x - e x + e - x d d x x 2 + 1 x 2 + 1 2 [Using chain rule of differentiation] = x 2 + 1 e x + e - x - 1 - e x + e - x 2 x x 2 + 1 2 ∴ d d x e x + e - x x 2 + 1 = x 2 + 1 e x - e - x - e x + e - x 2 x x 2 + 1 2 .

Differentiate : ln x x 2

We have to differentiate ln x x 2 . So, d d x ln x x 2 = x 2 d d x ln x - ln x d d x x 2 x 4 [Using Quotient rule of differentiation] = x 2 1 x - ln x 2 x x 4 ∴ d d x ln x x 2 = x - 2 x ln x x 4 .

Show that f x = x sin 1 x , x ≠ 0 0 , x = 0 is not differentiable at x = 0 .

Given, f x = x sin 1 x , x ≠ 0 0 , x = 0 at x = 0 , we have f ' x = lim x → 0 f x - f 0 x - 0 = lim x → 0 x sin 1 x - 0 x = lim x → 0 sin 1 x = sin 1 0 So, the limit does not exist. Hence, the function is not differentiable at 0 .