Question

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

Answer 1

Answer: -4

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