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Let $f(x) = \dfrac{x}{(x-1)(2x+3)}$, where $x>1$. Then the $4^{th}$ derivative of $f, \: f^{(4)} (x)$ is equal to

  1. $- \frac{24}{5} \bigg[ \frac{1}{(x-1)^5} – \frac{48}{(2x+3)^5} \bigg]$
  2. $\frac{24}{5} \bigg[ –  \frac{1}{(x-1)^5} + \frac{48}{(2x-3)^5} \bigg]$
  3. $\frac{24}{5} \bigg[ \frac{1}{(x-1)^5} + \frac{48}{(2x+3)^5} \bigg]$
  4. $\frac{64}{5} \bigg[ \frac{1}{(x-1)^5} + \frac{48}{(2x+3)^5} \bigg]$
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@`JEET

use partial fraction.

$\frac{x}{(x-1)(2x+3)} = \frac{A}{(x-1)} + \frac{B}{(2x+3)}$

$\Rightarrow x = A(2x+3) + B(x-1)$

when $x=1$, then $1=5A$ i.e. $A= 1/5 $

when $x=-3/2$, then $-3/2=-5B/2$ i.e. $B= 3/5 $

So, $f(x) = \frac{x}{(x-1)(2x+3)} = \frac{1}{5(x-1)} + \frac{3}{5(2x+3)}$

now, successive differentiation will be easy :)

1
Thanks.
0

@ankitgupta.1729

I am not getting this 48 in the differentiation of $\frac{3}{5(2x+3)}$.

Can you please verify once??

2

@`JEET

Here, $f(x) = \frac{1}{5(x-1)} + \frac{3}{5(2x+3)}\\ f'(x) =\frac{-1}{5(x-1)^2} - \frac{6}{5(2x+3)^2}\\ f''(x) = \frac{2}{5(x-1)^3} + \frac{24}{5(2x+3)^3}\\ f'''(x) = \frac{-6}{5(x-1)^4} - \frac{72*2}{5(2x+3)^4} \\f''''(x) = \frac{24}{5(x-1)^5} + \frac{72*2*8}{5(2x+3)^5}$

answer should be (C)

1
I made a foolish mistake. I wasn't differentiating the $2x$ itself.

Between thanks.

1 Answer

1 vote

Answer: $\mathbf C$

Explanation:

Using partial differentiation:

$\frac{x}{(x-1)(2x+3)} = \frac{\mathrm A}{x-1} + \frac{\mathrm B}{(2x+3)}$

$\Rightarrow \mathrm x = \mathrm A(2x+3) + \mathrm B(x-1)$

When $x = 1$, then $1 = 5\mathrm A \Rightarrow \mathrm A = \frac{1}{5}$

when $x = -\frac{3}{2}, \text {then}\; -\frac{3}{2} = -\frac{5\mathrm B}{2} \Rightarrow \mathrm B = \frac{3}{5}$

So, $f(x) = \frac{x}{(x-1)(2x+3)} = \frac{1}{(x-1)}+\frac{3}{5(3x+3)}$

Now, $f(x) = \frac{1}{5(x-1)} + \frac{3}{5(2x+3)}\\ f'(x) =\frac{-1}{5(x-1)^2} - \frac{6}{5(2x+3)^2}\\ f''(x) = \frac{2}{5(x-1)^3}+ + \frac{24}{5(2x+3)^3}\\ f'''(x) = \frac{-6}{5(x-1)^4} - \frac{72\times2}{5(2x+3)^4} \\f''''(x) = \frac{24}{5(x-1)^5} + \frac{72\times2\times8}{5(2x+3)^5} \\f''''(x) = \frac{24}{5}\bigg[\frac{1}{(x-1)^5} + \frac{48}{5(2x+3)^5}\bigg ]$

$\therefore \mathrm {\mathbf C}$ is the correct option.


edited by
1
Can you make another line to show the closed form like those in the options?

I meant to show
$=\frac{24}{5} \bigg[ \frac{1}{(x-1)^5} + \frac{48}{(2x+3)^5} \bigg]$ being the last line.
1
Done!

Thanks for pointing out.

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