If x^3 + 8xy + y^3 = 64, then dy dx = - Vidyadip

MCQ

If ${x^3} + 8xy + {y^3} = 64$, then ${{dy} \over {dx}} = $

✓
$ - {{3{x^2} + 8y} \over {8x + 3{y^2}}}$
B
${{3{x^2} + 8y} \over {8x + 3{y^2}}}$
C
${{3x + 8{y^2}} \over {8{x^2} + 3y}}$
D
None of these

✓

Answer

Correct option: A.

$ - {{3{x^2} + 8y} \over {8x + 3{y^2}}}$

a
(a) ${x^3} + 8xy + {y^3} = 64$

$ \Rightarrow 3{x^2} + 8\left( {y + x\frac{{dy}}{{dx}}} \right) + 3{y^2}\frac{{dy}}{{dx}} = 0$

$\therefore \frac{{dy}}{{dx}} = - \frac{{3{x^2} + 8y}}{{8x + 3{y^2}}}$.

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