If $x^{2}+y^{2}=1$, then $\frac{d^{2} x}{d y^{2}}=$

Question

Aicharya · Accepted Answer

We have $x^{2}+y^{2}=1$ $\begin{array}{l} 2 x+2 y \frac{d y}{d x}=0 \Rightarrow 2 y \frac{d y}{d x}=-2 x \ \frac{d y}{d x}=-\frac{x}{y} \Rightarrow \frac{d x}{d y}=-\frac{y}{x} \end{array}$ Differentiating w.r.t. $\mathrm{y}$ we get $\frac{d^{2} x}{d y^{2}}=\left[\frac{x(1)-y \frac{d x}{d y}}{x^{2}}ight]=-\left[\frac{x-y\left(\frac{-y}{x}ight)}{x^{2}}ight]=-\left[\frac{x^{2}+y^{2}}{x^{3}}ight]$ $=-\frac{1}{x^{3}} \quad \cdots\left[\because x^{2}+y^{2}=1ight.$, given $]$

If $x^{2}+y^{2}=1$, then $\frac{d^{2} x}{d y^{2}}=$

Solution