MCQ
જો ${x^3} + 8xy + {y^3} = 64$, તો ${{dy} \over {dx}} = $
- ✓$ - {{3{x^2} + 8y} \over {8x + 3{y^2}}}$
- B${{3{x^2} + 8y} \over {8x + 3{y^2}}}$
- C${{3x + 8{y^2}} \over {8{x^2} + 3y}}$
- Dએકપણ નહીં
$ \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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