UPSC NDA Math Model Paper 2

Given: $y=e^{x}$ To Find: $\left(\frac{d^{2}x}{dy^{2}}\right) \times \left(\frac{d^{2}y}{dx^{2}}\right)$ $y=e^{x}$ Differentiating with respect to x, we get $\Rightarrow \frac{dy}{dx} = e^{x}$ Again differentiating with respect to x, we get $\Rightarrow \frac{d^{2}y}{dx^{2}} = e^{x}$ ......(1) Again, $y=e^{x}$ Taking log both sides, we get $\Rightarrow \log y = \log \Rightarrow e^{x}$ $\Rightarrow \log y = x \log e$ $(\because \log m^{n} = n \log m)$ $\Rightarrow \log y = x$ $(\because \log e = 1)$ $\Rightarrow x = \log y$ Differentiating with respect to y, we get $\Rightarrow \frac{dx}{dy} = \frac{1}{y}$ Again differentiating with respect to y, we get $\Rightarrow \frac{d^{2}x}{dy^{2}} = \frac{-1}{y^{2}}$ .......(2) Multiplying equation (1) and (2), we get $\left(\frac{d^{2}x}{dy^{2}}\right) \times \left(\frac{d^{2}y}{dx^{2}}\right) = \frac{-1}{y^{2}} \times e^{x}$ $\Rightarrow \left(\frac{d^{2}x}{dy^{2}}\right) \times \left(\frac{d^{2}y}{dx^{2}}\right) = \frac{-1}{y^{2}} \times y$ $(\because y = e^{x})$ $\therefore \left(\frac{d^{2}x}{dy^{2}}\right) \times \left(\frac{d^{2}y}{dx^{2}}\right) = \frac{-1}{y}$