Question

If $x^{3}+y^{3}=t+\frac{4}{t}$ and $x^{6}+y^{6}=t^{2}+\frac{16}{t^{2}}$, then find $x^{4} y^{2} \frac{d y}{d x}$

Answer 1

$x^{3}+y^{3}=t+\frac{4}{t}$
$x^{6}+y^{6}+2 x^{3} y^{3}=t^{2}+\frac{16}{t^{2}}+8$ ...[By squaring both the sides]
$\Rightarrow\left(t^{2}+\frac{16}{t^{2}}\right)+2 x^{3} y^{3}=t^{2}+\frac{16}{t^{2}}+8$
$\Rightarrow x^{3} y^{3}=4$ ...(i)
Differentiating with respect to $x$, we get
$x^{3}\left(3 y^{2} \frac{d y}{d x}\right)+y^{3} \cdot\left(3 x^{2}\right)=0$
$\Rightarrow 3 x^{3} y^{2} \frac{d y}{d x}=-3 x^{2} y^{3}$
$\Rightarrow x^{3} y^{2} \frac{d y}{d x}=-x^{2} y^{3}$
$\Rightarrow x^{4} y^{2} \frac{d y}{d x}=-x^{3} y^{3}$
$\Rightarrow x^{4} y^{2} \frac{d y}{d x}=-4$ ...[From (i)]