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Explain solution RD Sharma class 12 Chapter 11 Higher Order Derivatives Exercise Multiple Choice Question question 23

Answers (1)

Answer:

                (c)

Hint:

We must know about the rules of finding the derivative.

Given:

                \frac{d}{d x}\left\{x^{n}-a_{1} x^{n-1}+a_{2} x^{n-2}+\ldots .+(-1)^{n} a_{n}\right\} e^{x}=x^{n} e^{x} \\

Explanation:

                \frac{d}{d x}\left\{x^{n}-a_{1} x^{n-1}+a_{2} x^{n-2}+\ldots .+(-1)^{n} a_{n}\right\} e^{x}=x^{n} e^{x} \\

                \frac{d}{d x}\left[x^{n}-a_{1} x^{n-1}+a_{2} x^{n-2}+\ldots .+(-1)^{n} a_{n}\right] e^{x} \\

                \begin{aligned} & &\frac{d}{d x}\left[x^{n}-n x^{n-1}+n(n-1) x^{n-2}+\ldots . .+(-1)^{n} a_{n}\right] e^{x} \end{aligned}

Comparing the coefficients of above two equations,

                a_{1}=n, a_{2}=n(n-1)

Similarly,

                a_{r}=n(n-1)(n-2)(n-3) \ldots(n-r+1) \\

                \begin{aligned} & &a_{r}=\frac{n !}{(n-r) !} \end{aligned}

 

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