If $x^2+y^2=\mathrm{t}+\frac{1}{\mathrm{t}}, x^4+y^4=\mathrm{t}^2+\frac{1}{\mathrm{t}^2}$, then…
If $x^2+y^2=\mathrm{t}+\frac{1}{\mathrm{t}}, x^4+y^4=\mathrm{t}^2+\frac{1}{\mathrm{t}^2}$, then $\frac{\mathrm{d} y}{\mathrm{~d} x}=$
- $\frac{1}{x^3 y}$
- $\frac{1}{x y^3}$
- $-\frac{1}{x y^3}$
- $-\frac{1}{x^3 y}$
Solution
$\begin{array}{ll} & x^2+y^2=\mathrm{t}+\frac{1}{\mathrm{t}} \text { and } x^4+y^4=\mathrm{t}^2+\frac{1}{\mathrm{t}^2} \\ \therefore \quad & \left(x^2+y^2\right)^2=\left(\mathrm{t}+\frac{1}{\mathrm{t}}\right)^2 \\ \therefore \quad & x^4+y^4+2 x^2 y^2=\mathrm{t}^2+\frac{1}{\mathrm{t}^2}+2 \\ \therefore \quad & \mathrm{t}^2+\frac{1}{\mathrm{t}^2}+2 x^2 y^2=\mathrm{t}^2+\frac{1}{\mathrm{t}^2}+2 \\ \therefore \quad & x^2 \dot{y}^2=1 \\ \therefore \quad & y^2=\frac{1}{x^2}\end{array}$
Differentiating w.r.t. $x$, we get
$\begin{aligned}
& 2 y \frac{\mathrm{~d} y}{\mathrm{~d} x}=\frac{-2}{x^3} \\
\therefore \quad & \frac{\mathrm{~d} y}{\mathrm{~d} x}=\frac{-1}{x^3 y}
\end{aligned}$
Asked in: MHT CET 2024 (11 May Shift 1)
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