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Question Number 188492 by 073 last updated on 02/Mar/23

Commented by Rasheed.Sindhi last updated on 05/Mar/23

$I s^{'} {d e r}^{'} m e a n s d e r i v a t i v e ?$
Commented by 073 last updated on 06/Mar/23

$der is degree of polynomal$
	Answered by Rasheed.Sindhi last updated on 05/Mar/23

	$d e r [P (x)] = 5 \Rightarrow P (x) = 5 x + C$ $d e r [Q (x)] = 4 \Rightarrow Q (x) = 4 x + C$ $\frac{P (x) \cdot Q (x)}{P (x) + Q (x)} = \frac{(5 x + C) (4 x + C)}{(5 x + C) + (4 x + C)}$ $= \frac{20 x^{2} + C}{9 x + C}$ $d e r (\frac{P (x) \cdot Q (x)}{P (x) + Q (x)}) = d e r (\frac{20 x^{2} + C}{9 x + C})$ $= \frac{(9 x) d e r (20 x^{2}) - (20 x^{2}) d e r (9 x)}{{(9 x)}^{2}}$ $= \frac{(9 x) (40 x) - (20 x^{2}) (9)}{81 x^{2}}$ $= \frac{360 x^{2} - 180 x^{2}}{81 x^{2}}$ $= \frac{180 x^{2}}{81 x^{2}} = \frac{20}{9}$
	Commented by Rasheed.Sindhi last updated on 05/Mar/23

	$I^{'} v e c o r r e c t e d m y a n s w e r . P l e a s e$ $c o m m e n t w h e t h e r a n s w e r i s^{'} {o k}^{'}$ $o r^{'} n o t {o k}^{'} .$
	Commented by manxsol last updated on 05/Mar/23

	$C_{1 a n d} C_{2}$ $n o t C a n d C$
	Commented by 073 last updated on 06/Mar/23

	$no sir$
	Commented by Rasheed.Sindhi last updated on 06/Mar/23

	$A b o v e a n s w e r i s i n c a s e^{'} {d e r}^{'} m e a n s$ $d e r i v a t i v e .$
	Commented by Rasheed.Sindhi last updated on 06/Mar/23

	$@ m a n x s o l$ $s i r W i l l^{'} C_{1} & C_{2} i n s t e a d o f C^{'} a f f e c t$ $t h e r e s u l t ?$
	Commented by manxsol last updated on 06/Mar/23

	$e x e r c i s e i n i t i a l i s e a s y$ $d e r (\frac{x^{5} . x^{4}}{x^{5} + x^{4}}) =$ $d e r (\frac{x^{9}}{x^{5} + x^{4}}) = 9 - 5 = 4$ $y o u$ ${[\frac{P Q}{P + Q}]}^{'} =$ $\frac{(P + Q) [P^{'} Q + {P Q}^{'}] - [P^{'} + Q^{'}) P Q}{{(P + Q)}^{2}}$ $\frac{\overset{}{P P}^{'} Q + P^{2} Q^{'} + Q^{2} {\overset{}{P^{'}}}^{'} + \overset{}{{Q Q}^{'} P} - {P P}^{'} Q - {\overset{}{P Q Q}}^{'}}{{(P + Q)}^{2}} \frac{}{}$ $\frac{P^{2} Q^{'} + Q^{2} P^{'}}{{(P + Q)}^{2}}$ $P = 5 x + C_{1}$ $Q = 4 x + C_{2}$ $\frac{{(5 x + C_{1})}^{2} (4) + {(4 x + C_{2})}^{2} 5}{{(9 x + C_{1} + C_{2})}^{2}}$ $w$ $\frac{180 x^{2} + 40 x (C_{1} + C_{2}) + 4 C_{1}^{2} + 5 C_{2}^{2}}{81 x^{2} + 18 x (C_{1} + C_{2}) + (C_{1}^{2} + C_{2}^{2})}$ $s i r W i l l^{'} C_{1} & C_{2} i n s t e a d o f C^{'} a f f e c t$ $t h e r e s u l t ?$
	Answered by Rasheed.Sindhi last updated on 06/Mar/23

	$\underset{―}{I f^{'} {d e r}^{'} m e a n s d e g r e e o f p o l y n o m i a l ._{}}$ $d e r [P (x)] = 5, d e r [Q (x)] = 4$ $d e r [\frac{P (x) \cdot Q (x)}{P (x) + Q (x)}] = ?$ $N o t e t h a t \frac{P (x) \cdot Q (x)}{P (x) + Q (x)} i s p o l y n o m i a l$ $i f [P (x) + Q (x)] ∣ P (x) \cdot Q (x)$ $A s s u m i n g a b o v e c o n d i t i o n :$ $∙ N = d e r [P (x) . Q (x)] = d e r [P (x)] + d e r [Q (x)]$ $= 5 + 4 = 9$ $∙ D = d e r [P (x) + Q (x)] = m a x (d e r [P (x)] + d e r [Q (x)])$ $= m a x (5, 4) = 5$ $∙ d e r [\frac{N}{D}] = d e r (N) - d e r (D)$ $= 9 - 5 = 4$
	Commented by 073 last updated on 06/Mar/23

	$nice solution sir$
	Commented by Rasheed.Sindhi last updated on 06/Mar/23

	$T h a n k y o u f o r r a p i d r e s p o n s e .$
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