To show that a function is differentiable at x = 0, we need to check if the function is continuous at x = 0 and if the derivative exists at x = 0. (a) For the function f(x) = |x| sin x, we can see that it is continuous at x = 0 because both |x| and sin x are continuous functions. To check if the derivative exists at x = 0, we need to calculate the left and right-hand derivatives separately. For x 0, f(x) = x sin x. Taking the derivative of f(x) with respect to x, we get f'(x) = sin x + x cos x. As x approaches 0 from the right, the derivative becomes f'(0+) = sin 0 + 0 cos 0 = 0. Since the left and right-hand derivatives are both 0, the derivative of f(x) exists at x = 0. Therefore, f(x) = |x| sin x is differentiable at x = 0. (b) For the function f(x) = x^(2/3) sin x, we can see that it is continuous at x = 0 because both x^(2/3) and sin x are continuous functions. To check if the derivative exists at x = 0, we need to calculate the left and right-hand derivatives separately. For x 0, f(x) = x^(2/3) sin x. Taking the derivative of f(x) with respect to x, we get f'(x) = 2/3 x^(-1/3) sin x + x^(2/3) cos x.

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To show that a function is differentiable at x = 0, we need to check if the function is continuous at x = 0 and if the derivative exists at x = 0. (a) For the function f(x) = |x| sin x, we can see that it is continuous at x = 0 because both |x| and sin x are continuous functions. To check if the derivative exists at x = 0, we need to calculate the left and right-hand derivatives separately. For x < 0, f(x) = -x sin x. Taking the derivative of f(x) with respect to x, we get f'(x) = -sin x - x cos x. As x approaches 0 from the left, the derivative becomes f'(0-) = -sin 0 - 0 cos 0 = 0. For x > 0, f(x) = x sin x. Taking the derivative of f(x) with respect to x, we get f'(x) = sin x + x cos x. As x approaches 0 from the right, the derivative becomes f'(0+) = sin 0 + 0 cos 0 = 0. Since the left and right-hand derivatives are both 0, the derivative of f(x) exists at x = 0. Therefore, f(x) = |x| sin x is differentiable at x = 0. (b) For the function f(x) = x^(2/3) sin x, we can see that it is continuous at x = 0 because both x^(2/3) and sin x are continuous functions. To check if the derivative exists at x = 0, we need to calculate the left and right-hand derivatives separately. For x < 0, f(x) = -(-x)^(2/3) sin (-x) = -x^(2/3) sin x. Taking the derivative of f(x) with respect to x, we get f'(x) = -2/3 x^(-1/3) sin x - x^(2/3) cos x. As x approaches 0 from the left, the derivative becomes f'(0-) = -2/3 (0)^(-1/3) sin 0 - (0)^(2/3) cos 0 = 0. For x > 0, f(x) = x^(2/3) sin x. Taking the derivative of f(x) with respect to x, we get f'(x) = 2/3 x^(-1/3) sin x + x^(2/3) cos x.

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To show that a function is differentiable at x = 0, we need to check if the function is continuous at x = 0 and if the derivative exists at x = 0. (a) For the function f(x) = |x| sin x, we can see that it is continuous at x = 0 because both |x| and sin x are continuous functions. To check if the derivative exists at x = 0, we need to calculate the left and right-hand derivatives separately. For x < 0, f(x) = -x sin x. Taking the derivative of f(x) with respect to x, we get f'(x) = -sin x - x cos x. As x approaches 0 from the left, the derivative becomes f'(0-) = -sin 0 - 0 cos 0 = 0. For x > 0, f(x) = x sin x. Taking the derivative of f(x) with respect to x, we get f'(x) = sin x + x cos x. As x approaches 0 from the right, the derivative becomes f'(0+) = sin 0 + 0 cos 0 = 0. Since the left and right-hand derivatives are both 0, the derivative of f(x) exists at x = 0. Therefore, f(x) = |x| sin x is differentiable at x = 0. (b) For the function f(x) = x^(2/3) sin x, we can see that it is continuous at x = 0 because both x^(2/3) and sin x are continuous functions. To check if the derivative exists at x = 0, we need to calculate the left and right-hand derivatives separately. For x < 0, f(x) = -(-x)^(2/3) sin (-x) = -x^(2/3) sin x. Taking the derivative of f(x) with respect to x, we get f'(x) = -2/3 x^(-1/3) sin x - x^(2/3) cos x. As x approaches 0 from the left, the derivative becomes f'(0-) = -2/3 (0)^(-1/3) sin 0 - (0)^(2/3) cos 0 = 0. For x > 0, f(x) = x^(2/3) sin x. Taking the derivative of f(x) with respect to x, we get f'(x) = 2/3 x^(-1/3) sin x + x^(2/3) cos x.

Show that the following functions are diﬀerentiable at x = 0.(a) |x| sin x (b) x2/3 sin x (c) 3√x(1−cos x) (d) h(x) ={x2 sin(1/x), x 6 = 00, x = 0.Page 3

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