Show that the following functions are differentiable at x = 0.(a) |x| sin x

To show that the function |x| sin x is differentiable at x = 0, we need to check if the limit of the difference quotient exists as x approaches 0.

Let's start by finding the derivative of |x| sin x. The derivative of |x| is not defined at x = 0, but we can use the definition of the derivative to find the derivative of |x| sin x for x ≠ 0.

For x ≠ 0, the derivative of |x| sin x can be found using the product rule. The derivative of |x| is sign(x) and the derivative of sin x is cos x. So, the derivative of |x| sin x is given by:

d/dx (|x| sin x) = sign(x) sin x + |x| cos x

Now, let's evaluate the limit of the difference quotient as x approaches 0. The difference quotient is given by:

lim (h -> 0) [(|h| sin h - 0) / h]

Since |h| sin h is a continuous function, we can directly substitute h = 0 into the expression:

lim (h -> 0) [(|h| sin h - 0) / h] = lim (h -> 0) [(0 - 0) / h] = lim (h -> 0) [0 / h] = lim (h -> 0) 0 = 0

Since the limit of the difference quotient is 0, the function |x| sin x is differentiable at x = 0.

Therefore, we have shown that the function |x| sin x is differentiable at x = 0.