Please solve RD Sharma class 12 chapter 11 Higher Order Derivatives exercise very short answer type question 5 maths textbook solution

\begin{aligned} &\text { The value of }\frac{d^{2} y}{d x^{2}}=\frac{f^{\prime} g^{\prime \prime}-g^{\prime} f^{\prime \prime}}{f^{13}} \end{aligned}

Hint:

Diff $x$  and $y$ w.r.t  $t$

$\text { Use } \frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}} \text { again differentiate it to get } \frac{d^{2} y}{d x^{2}}$

Given:

$x=f(t)\: and\: y=g(t)$

Explanation:

It is given that

$x=f(t)\: and\: y=g(t)$

$\frac{dx}{dt}=f'(t)\: and\: \frac{dy}{dt}=g'(t)$

So,

\begin{aligned} &\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}} \\ &\frac{d y}{d x}=\frac{g^{\prime}(t)}{f^{\prime}(t)} \end{aligned}

Diff. w.r.t $x$

\begin{aligned} \frac{d}{d x}\left(\frac{d y}{d x}\right) &=\frac{d}{d x}\left(\frac{g^{\prime}(t)}{f^{\prime}(t)}\right) \\ &=\frac{d}{d t}\left(\frac{g^{\prime}(t)}{f^{\prime}(t)}\right) \cdot \frac{d t}{d x} \\ &=\frac{f^{\prime} \cdot g^{\prime \prime}-g^{\prime} f^{\prime \prime}}{\left(f^{\prime}\right)^{2}} \cdot\left(\frac{1}{f^{\prime}(t)}\right) \\ \frac{d^{2} y}{d x^{2}}=\frac{f^{\prime} g^{\prime \prime}-g^{\prime} f^{\prime \prime}}{f^{13}} \end{aligned}

\begin{aligned} &\text { Thus the value of }\frac{d^{2} y}{d x^{2}}=\frac{f^{\prime} g^{\prime \prime}-g^{\prime} f^{\prime \prime}}{f^{13}} \end{aligned}