15) Its given \( y=x^{2} e^{a x} \) then \( y^{(n)} \) a) \( e^{a x}\left(a^{n} x^{2}-2 n a^{n-1} x+n(n-1) a^{n-2}\right) \) b) \( e^{a x}\left(a^{n} x^{2}+2 n a^{n-1} x-n(n-1) a^{n-2}\right) \) c) \( e^{a x}\left(a^{n} x^{2}-2 n a^{n-1} x-n(n-1) a^{n-2}\right) \) d) \( e^{a x}\left(a^{n} x^{2}+2 n a^{n-1} x+n(n-1) a^{n-2}\right) \) e) None of these

To find the \( n \)-th derivative \( y^{(n)} \) of the function \( y = x^2 e^{ax} \), we can use the product rule and Faá di Bruno's formula for higher derivatives. The key points include recognizing the behavior of \( e^{ax} \) and applying the Leibniz rule for derivatives of products. You'll find that, ultimately, the expression for \( y^{(n)} \) takes into account both the derivatives of \( x^2 \) and the exponential function, resulting in terms that involve powers of \( a \) and algebraic expressions dependent on \( n \). After careful application of these principles, you can derive that the correct option is: **a)** \( e^{a x}\left(a^{n} x^{2}-2 n a^{n-1} x+n(n-1) a^{n-2}\right) \)