Question

Where are the functions \(f_{1}(x)=|\sin x|\) and \(f_{2}(x)=\sin |x|\) differentiable?

Short answer

The function \(f_{1}(x)=|\sin x|\) is differentiable everywhere except at \(x = n\pi\) and the function \(f_{2}(x)=\sin |x|\) is differentiable everywhere except at \(x=0\).

Step-by-step solution

Differentiability of \(f_{1}(x)\)

Start by looking at the function \(f_{1}(x)=|\sin x|\). The function \(\sin x\) is differentiable for all real numbers. However, the absolute value function is not differentiable at zero. Therefore, the function \(f_{1}(x)=|\sin x|\) will not be differentiable whenever \(\sin x = 0\). Remembering the properties of sine function, it equals to zero at \(x = n\pi\) where \(n\) is an integer.

Differentiability of \(f_{2}(x)\)

Now, observe the function \(f_{2}(x)=\sin |x|\). The sine function is differentiable for all real numbers, so the issue here would be when \(|x|\) is not differentiable. Like in the first part of the problem, an absolute value function is not differentiable at zero. Therefore, \(f_{2}(x)=\sin |x|\) is not differentiable at \(x=0\).

Conclusion

Having analysed both functions, we can conclude that \(f_{1}(x)=|\sin x|\) is differentiable everywhere except at \(x = n\pi\), where \(n\) is an integer. The second function, \(f_{2}(x)=\sin |x|\) is differentiable everywhere except at \(x=0\).