Solveeit Logo
Login

Question

Question: If \(y = x^{2}\log x,\) then value of \(y_{n}\) is...

If y=x2log⁡x,y = x^{2}\log x, then value of yny_{n} is

A

(−1)n−1(n−3)!xn−2\frac{( - 1)^{n - 1}(n - 3)!}{x^{n - 2}}

B

(−1)n−1(n−3)!xn−2.2\frac{( - 1)^{n - 1}(n - 3)!}{x^{n - 2}}.2

C

(−1)n−1(n−2)!xn−2\frac{( - 1)^{n - 1}(n - 2)!}{x^{n - 2}}

D

None of these

Answer

(−1)n−1(n−3)!xn−2.2\frac{( - 1)^{n - 1}(n - 3)!}{x^{n - 2}}.2

Explanation

Solution

Applying Leibnitz’s theorem by taking x2x^{2} as second function, we get, Dny=Dn(log⁡x.x2)D^{n}y = D^{n}(\log x.x^{2})

= nC0Dn(log⁡x).x2+n⥂C1Dn−1(log⁡x).D(x2)+n⥂C2Dn−2(log⁡x)D2(x2)+...........nC_{0}D^{n}(\log x).x^{2} +^{n} ⥂ C_{1}D^{n - 1}(\log x).D(x^{2}) +^{n} ⥂ C_{2}D^{n - 2}(\log x)D^{2}(x^{2}) + ........... = (−1)n−1(n−1)!xn.x2+n.(−1)n−2(n−2)!xn−1.2x+n(n−1)2!(−1)n−3(n−3)!xn−2.2+0+0.........\frac{( - 1)^{n - 1}(n - 1)!}{x^{n}}.x^{2} + n.\frac{( - 1)^{n - 2}(n - 2)!}{x^{n - 1}}.2x + \frac{n(n - 1)}{2!}\frac{( - 1)^{n - 3}(n - 3)!}{x^{n - 2}}. 2 + 0 + 0......... = (−1)n−1(n−1)!xn−2+2n(−1)n−2(n−2)!xn−2+n(n−1)(−1)n−3(n−3)!xn−2\frac{( - 1)^{n - 1}(n - 1)!}{x^{n - 2}} + \frac{2n( - 1)^{n - 2}(n - 2)!}{x^{n - 2}} + \frac{n(n - 1)( - 1)^{n - 3}(n - 3)!}{x^{n - 2}}

= (−1)n−1(n−3)!xn−2×{(n−1)(n−2)−2n(n−2)+n(n−1)}\frac{( - 1)^{n - 1}(n - 3)!}{x^{n - 2}} \times \{(n - 1)(n - 2) - 2n(n - 2) + n(n - 1)\}

=. (−1)n−1(n−3)!xn−2.2\frac{( - 1)^{n - 1}(n - 3)!}{x^{n - 2}}.2