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Q.
If $y = e^{nx} , $ then $\left(\frac{d^{2}y}{dx^{2}}\right) \left(\frac{d^{2}x}{dy^{2}}\right) $ is equal to:
Continuity and Differentiability
Solution:
Given that, $ y = e^{nx}$
Differeniating both sides with respect to x
$\frac{dy}{dx} = ne^{nx}$
Again differentiating w.r.t 'x',
$ \frac{d^{2}y}{dx^{2}} = n.ne^{nx} =n^{2}e^{nx}$ .....(1)
$ y = e^{nx} $
$nx = \log_{e}y $
$x = \frac{1}{n} \log y$
Differentiating x with respect to y
$ \frac{dx}{dy} = \frac{1}{n}. \frac{1}{y} $
Again differentiating with respect to y
$\frac{d^{2}x}{dy^{2} } = \frac{1}{n}.\left(\frac{-1}{y^{2}}\right) = \frac{-1}{ny^{2}}= - \frac{1}{ne^{2nx}}$ ......(2)
Multiplying equation (1) and (2)
$ \left(\frac{d^{2}y}{dx^{2}}\right)\left(\frac{d^{2}x}{dy^{2}}\right) = \left(n^{2}e^{nx}\right) \left( \frac{-1}{n} e^{2nx}\right) = - ne^{-nx} $