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Find nth order derivative of \[y= \dfrac{x^2+4}{(2x+3)(x-1)^2}\]

written 4.6 years ago by teamques10 ★ 70k

modified 4.1 years ago by pg1118329 • 0

engineering mathematics

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written 4.6 years ago by teamques10 ★ 70k

We have

$y=\dfrac {x^2 + 4}{(2x+3)(x-1)^2}$

$ \therefore y = \dfrac {-1}{2x+3} + \dfrac {1}{(x-1)^2} $

Using formula if $ y=\dfrac {1}{(ax-b)}$ then $y_n = \dfrac {(-1)^n n!a^n}{(ax+b)^{n+1}}$

$\therefore y_n = \dfrac {(-1)^n n!2^n}{(2x-3)^{n+1}} + \dfrac {(-1)^n (n+1)!} {(x-1)^{n+2}}$

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