Discussion Forum

If $\frac{dy}{dx}=4$  and $\frac{d^{2}y}{d^{2}x}=-3$ at a point P on the curve y=f(x), then 

$\left(\frac{d^{2}x}{dy^{2}}\right)=$

Answer:

Explanation:

$\frac{dx}{dy}=\left(\frac{dy}{dx}\right)^{-1}$

  $\frac{d^{2}x}{dy^{2}}=\frac{-1}{\left(\frac{dy}{dx}\right)^{2}}\times\frac{d}{dx}\left(\frac{dy}{dx}\right)\times\left(\frac{dx}{dy}\right)$

 $\Rightarrow$     $\frac{d^{2}x}{dy^{2}}=\frac{-d^{2}y}{dx^{2}}/\left(\frac{dy}{dx}\right)^{3}$

$\left(\frac{d^{2}x}{dy^{2}}\right)_{P}=-\left(\frac{-3}{(4)^{3}}\right)=\frac{3}{64}$

                                                                                    $\left[\because \frac{dy}{dx}=4,\frac{d^{2}x}{dx^{2}}=-3\right]$