Right option is (b) \(\frac{3 sin a^3 – 2 sin a^2}{a}\)
The explanation is: Applying the Leibniz rule equation given by
\( f’ (α) = \int_p^q \frac{∂}{∂α} f (x,α) dx + f (q, α) \frac{dq}{dα} – f (p, α) \frac{dp}{dα} …..(1)\)
\(f (x,a) = \frac{sinax}{x}, p=a, q=a^2\) & further obtaining
\(f(q,a) = f(a^2,a) = \frac{sin a^3}{a^2}, \frac{dq}{da} = 2a\)
\(f(p,a) = f(a,a) = \frac{sin a^2}{a}, \frac{dp}{da} = 1\)
substituting all these values in (1) we get
\(f’(a) = \int_a^{a^2} \frac{∂}{∂α} (\frac{sinax}{x})dx + \frac{sin a^3}{a^2}.2a – \frac{sin a^2}{a}.1\)
\(\int_a^{a^2} \frac{1}{x} (cos(ax))(x)+ \frac{2 sin a^3 – sin^2 a}{a}\)
\([\frac{sinax}{x}]_a^{a^2} + \frac{2 sin a^3 – sin^2 a}{a} = \frac{sin a^3}{a} – \frac{sin a^2}{a} + \frac{2 sin a^3 – sin^2 a}{a}\)
thus \(f’(a) = \frac{3 sin a^3 – 2 sin a^2}{a}.\)