解：(a) f(x)=d[sin(9x^2)/(9x^2)/dx=[18xcos(9x^2)9x^2-sin(9x^2)18x]/(81x^4)
=[162x^3cos(9x^2)-18xsin(9x^2)]/(81x^4)
=2cos(9x^2)/x-2sin(9x^2)/(9x^3).
(b)
f''(√π)={[-2*18xsin(9x^2)x-2cos(9x^2)]/x^2-[2*18xcos(9x^2)9x^3-2*27x^2sin(9x^2)]/(81x^6)}(√π)
=[-36sin(9x^2)-2cos(9x^2)/x^2-4cos(9x^2)/x^2+2sin(9x^2)/(3x^4)](π)
=-6cos(9π)/π=6/π
f'(x)=34sin9x^2/x
f''(x)=-612cos9x^2 -sin9x^2/x^2
所以
f''(π/2)=-612cos9π
所以f''(π/2)=612
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