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Question Number 36101 by rahul 19 last updated on 28/May/18

Commented by rahul 19 last updated on 29/May/18
Ans. given is 3.
Commented by tanmay.chaudhury50@gmail.com last updated on 29/May/18

$p l s c h e c k$
	Answered by tanmay.chaudhury50@gmail.com last updated on 29/May/18
	$y=f(x)=x^7 (dy/dx)=f_1 (x)=7x^6 (d^2 y/dx^2 )=f_2 (x)=42x^5 (d^3 y/dx^3 )=f_3 (x)=210x^4 (dx/dy)=(1/(f_1 (x)))=(1/(7x^6 )) (d^2 x/dy^2 )=(d/dy)((dx/dy)) =(d/dx){(1/(f_1 (x)))}(dx/dy) =((−1)/({f_1 (x)}^2 ))×f_2 (x)×(1/(f_1 (x)))=((−f_2 (x))/({f_1 (x)}^3 ))=((−42x^5 )/(343x^(18) ))=((−6)/(49x^(13) )) (d^3 x/dy^3 )=(d/dy)((d^2 x/dy^2 ))=(d/dx){((−f_2 (x))/((f_1 (x))^3 ))}×(dx/dy) =[((−{f_1 (x)}^3 ×f_3 (x)+f_2 (x)×3{f_1 (x)}^2 ×f_2 (x))/({f_1 (x)}^6 ))](1/(f_1 (x))) now putting the value =((−(7x^6 )^3 ×210x^4 +42x^5 ×3×49x^(12) ×42x^5 )/((7x^6 )^6 ))×(1/(7x^6 )) =((42×42×49×3×x^(22) −210×7^3 ×x^(22) )/(7^7 ×x^(42) )) =((7^4 ×3^3 ×2^2 x^(22) −7^4 ×5×3×2x^(22) )/(7^7 ×x^(42) )) =((7^4 ×x^(22) (108−30))/(7^7 ×x^(42) )) =((78)/(343x^(20) ))$
	$y = f (x) = x^{7}$ $\frac{d y}{d x} = f_{1} (x) = 7 x^{6}$ $\frac{d^{2} y}{{d x}^{2}} = f_{2} (x) = 42 x^{5}$ $\frac{d^{3} y}{{d x}^{3}} = f_{3} (x) = 210 x^{4}$ $\frac{d x}{d y} = \frac{1}{f_{1} (x)} = \frac{1}{7 x^{6}}$ $\frac{d^{2} x}{{d y}^{2}} = \frac{d}{d y} (\frac{d x}{d y})$ $= \frac{d}{d x} {\frac{1}{f_{1} (x)}} \frac{d x}{d y}$ $= \frac{- 1}{{f_{1} (x)}^{2}} \times f_{2} (x) \times \frac{1}{f_{1} (x)} = \frac{- f_{2} (x)}{{f_{1} (x)}^{3}} = \frac{- 42 x^{5}}{343 x^{18}} = \frac{- 6}{49 x^{13}}$ $\frac{d^{3} x}{{d y}^{3}} = \frac{d}{d y} (\frac{d^{2} x}{{d y}^{2}}) = \frac{d}{d x} {\frac{- f_{2} (x)}{{(f_{1} (x))}^{3}}} \times \frac{d x}{d y}$ $= [\frac{- {f_{1} (x)}^{3} \times f_{3} (x) + f_{2} (x) \times 3 {f_{1} (x)}^{2} \times f_{2} (x)}{{f_{1} (x)}^{6}}] \frac{1}{f_{1} (x)}$ $n o w p u t t i n g t h e v a l u e$ $= \frac{- {(7 x^{6})}^{3} \times 210 x^{4} + 42 x^{5} \times 3 \times 49 x^{12} \times 42 x^{5}}{{(7 x^{6})}^{6}} \times \frac{1}{7 x^{6}}$ $= \frac{42 \times 42 \times 49 \times 3 \times x^{22} - 210 \times 7^{3} \times x^{22}}{7^{7} \times x^{42}}$ $= \frac{7^{4} \times 3^{3} \times 2^{2} x^{22} - 7^{4} \times 5 \times 3 \times 2 x^{22}}{7^{7} \times x^{42}}$ $= \frac{7^{4} \times x^{22} (108 - 30)}{7^{7} \times x^{42}}$ $= \frac{78}{343 x^{20}}$
	Commented by rahul 19 last updated on 30/May/18
	sir, why 3 is not coming as the final answer . This expression is true for all x,y. Isn't it ?
	Commented by rahul 19 last updated on 31/May/18
	??��
	Answered by ajfour last updated on 29/May/18

	$\frac{d^{3} x}{{d y}^{3}} = \frac{d}{d y} [\frac{d}{d y} {(\frac{d y}{d x})}^{- 1}] = \frac{d}{d y} [- {(\frac{d y}{d x})}^{- 2} \frac{d^{2} y}{{d x}^{2}} (\frac{d x}{d y})]$ $\frac{d^{3} x}{{d y}^{2}} = \frac{d}{d x} [- {(\frac{d y}{d x})}^{- 3} \frac{d^{2} y}{{d x}^{2}}] (\frac{d x}{d y})$ $= 3 (\frac{d^{2} y}{{d x}^{2}}) {(\frac{d y}{d x})}^{- 5} - {(\frac{d y}{d x})}^{- 4} \frac{d^{3} y}{{d x}^{3}}$ $\Rightarrow {(\frac{d y}{d x})}^{7} \frac{d^{3} x}{{d y}^{3}} + {(\frac{d y}{d x})}^{3} \frac{d^{2} y}{{d x}^{2}} = 3 {(\frac{d y}{d x})}^{2} \frac{d^{2} y}{{d x}^{2}}$ $o r \frac{{(\frac{d y}{d x})}^{7} (\frac{d^{3} x}{{d y}^{3}}) + {(\frac{d y}{d x})}^{3} \frac{d^{2} y}{{d x}^{2}}}{{(\frac{d y}{d x})}^{2} \frac{d^{2} y}{{d x}^{2}}} = 3 .$
	Commented by tanmay.chaudhury50@gmail.com last updated on 29/May/18

	$e x c e l l e n t . . .$
	Commented by ajfour last updated on 29/May/18

	$m i s t a k e i n q u e s t i o n, i n t h e$ $d e n o m i n a t o r w e s h o u l d h a v e$ ${(\frac{d y}{d x})}^{2} \frac{d^{2} y}{{d x}^{2}} .$
	Commented by rahul 19 last updated on 30/May/18
	Thank you sir .
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