Let $ f(x)=x|x|, g(x)=\sin x $ and $ h(x)=(g o f)(x) $. Then

Question

Let $ f(x)=x|x|, g(x)=\sin x $ and $ h(x)=(g o f)(x) $. Then, $ h(x) $ is differentiable at $ x=0 $, but $ h^{\prime}(x) $ is not continuous at $ x=0 $, $ h(x) $ is not differentiable at $ x=0 $, $ h^{\prime}(x) $ is differentiable at $ x=0 $, $ h^{\prime}(x) $ is continuous at $ x=0 $ but is not differentiable at $ x=0 $

MARKS App · Accepted Answer

h x = sin ⁡ x 2 x ≥ 0 - sin ⁡ x 2 x 0 h ' 0 + = lim x → 0 + ⁡ sin ⁡ x 2 x = 0 h ' 0 - = lim x → 0 - ⁡ - sin ⁡ x 2 x = 0 Hence, h ' 0 = 0 So h ' x 2 x cos ⁡ x 2 x > 0 0 x = 0 - 2 x cos ⁡ x 2 x 0 Now, h '' 0 + = lim x → 0 + 2 x cos x 2 x = 2 h '' (0 - ) = lim x → 0 - ⁡ - 2 x cos ⁡ x 2 x = - 2 Hence, h ' x is non-derivable at x = 0 .

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