If 3 x 2-4 xy -2 y 2- x -8-0 then the value of dy - dx at -1-3 isA- 8 - 7B- 7 - 8C- -7 - 8D- -8 - 7

The correct option is C 7/8 3x^2+4xy+2y^2+x 8=0 ⇒ 6x+4(y+xy^1)+4y.y^1+1=0 ⇒ y^1= (6x+4y+1)/4(x+y) ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Dividing a Polynomial by a Polynomial

If 3 x 2+4 xy...

Question

$I f 3 x^{2} + 4 x y + 2 y^{2} + x - 8 = 0 t h e n t h e v a l u e o f \frac{d y}{d x} a t (- 1, 3) i s$

-7/8

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

7/8

No worries! We‘ve got your back. Try BYJU‘S free classes today!

-8/7

No worries! We‘ve got your back. Try BYJU‘S free classes today!

8/7

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

Suggest Corrections

Similar questions

Question 14
The multiplicative inverse of $- 1 \frac{1}{7}$ is
(a) $\frac{8}{7}$
(b) $\frac{- 8}{7}$
(c) $\frac{7}{8}$
(d) $\frac{- 7}{8}$

At $25^{\circ}$ C the pH of a $10^{- 8}$ molar solution of HCl in water is

8 x + 7 = 0

\Rightarrow 8 x = - 7 + 0

\Rightarrow 8 x = - 7

\Rightarrow x = 8 + 7

8 x + 7 = 0

The pH of $10^{- 8}$ M HCl solution is

\frac{d^{3} y}{d x^{3}} - 8 \frac{d^{2} y}{d x^{2}} = 0

y (0) = \frac{1}{8}, y^{1} (0) = 0

y^{2} (0) = 1

y^{n} (x) = \frac{d^{n} y}{d x^{n}}]

Join BYJU'S Learning Program