f(x)=∞∏k=1⎛⎜⎝1+2 cos(2x3k)3⎞⎟⎠=∞∏k=113(1+2−4sin2x3k)=∞∏k=113 sin(x3k){3 sin(x3k)−4 sin3(x3k)}=∞∏k=1⎡⎢⎣sin(x3k−1)3 sin(x3k)⎤⎥⎦=limk→∞13k⎧⎪⎨⎪⎩sin xsinx3×sinx3sinx9×.....×sin(x3k−1)sin(x3k)⎫⎪⎬⎪⎭=limk→∞sin xxsin(x3k)(x3k)⇒f(x)=sin xxxf(x)=sin x
So [sin x]+|sin x|+(x−1)|(x−1)(x−2)|
is non-differentiable at 5 (π2,π,2π,5π2,2) points