Question

If x = f(t) and y = g(t), then $\frac{d^{2}y}{d{x}^2}$is equal to

(a) $\frac{f\text{'}g\text{'}\text{'}-g\text{'}f\text{'}\text{'}}{{\left(f\text{'}\right)}^3}$
(b) $\frac{f\text{'}g\text{'}\text{'}-g\text{'}f\text{'}\text{'}}{{\left(f\text{'}\right)}^2}$
(c) $\frac{g\text{'}\text{'}}{f\text{'}\text{'}}$
(d) $\frac{f\text{'}\text{'}g\text{'}-g\text{'}\text{'}f\text{'}}{{\left(g\text{'}\right)}^3}$

Answer 1

(a) $\frac{f\text{'}g\text{'}\text{'}-g\text{'}f\text{'}\text{'}}{{\left(f\text{'}\right)}^3}$

Here,
x = f(t) and y = g(t)
$$\begin{gathered} \Rightarrow \frac{dx}{dt}=f\text{'}\left(t\right)\mathrm{and}\frac{dy}{dt}=g\mathit{\text{'}}\left(t\right) \\ \mathit{\therefore }\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}\mathit{=}\frac{\mathit{g}\mathit{\text{'}}\left(\mathit{t}\right)}{\mathit{f}\mathit{\text{'}}\left(\mathit{t}\right)} \end{gathered}$$


$$\begin{gathered} \Rightarrow \frac{{\mathit{d}}^{\mathit{2}}y}{\mathit{d}{x}^{\mathit{2}}}=\frac{d}{dt}\left\{\frac{g^{\text{'}}\left(t\right)}{f^{\text{'}}\left(t\right)}\right\}\times \frac{dt}{dx} \\ =\frac{f^{\text{'}}\left(t\right)g^{\text{'}\text{'}}\left(t\right)-g^{\text{'}}\left(t\right)f^{\text{'}\text{'}}\left(t\right)}{{\left[f^{\text{'}}\left(t\right)\right]}^2}\times \frac{1}{f\text{'}\left(t\right)} \\ =\frac{f^{\text{'}}\left(t\right)g^{\text{'}\text{'}}\left(t\right)-g^{\text{'}}\left(t\right)f^{\text{'}\text{'}}\left(t\right)}{{\left[f^{\text{'}}\left(t\right)\right]}^3} \end{gathered}$$