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

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

Answers (1)

Answer:

\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}

Posted by

Gurleen Kaur

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