Question

If $x={\left(t+\frac{1}{t}\right)}^a,y=a^{t+\frac{1}{t}},\mathrm{find}\frac{dy}{dx}$

Answer 1

$\mathrm{We}\mathrm{have},x={\left(t+\frac{1}{t}\right)}^a$

$$\begin{gathered} \Rightarrow \frac{dx}{dt}=\frac{d}{dt}\left[{\left(t+\frac{1}{t}\right)}^a\right] \\ \Rightarrow \frac{dx}{dt}=a{\left(t+\frac{1}{t}\right)}^{a-1}\frac{d}{dt}\left(t+\frac{1}{t}\right) \\ \Rightarrow \frac{dx}{dt}=a{\left(t+\frac{1}{t}\right)}^{a-1}\left(1-\frac{1}{t^2}\right)...\left(i\right) \\ \mathrm{and}, \\ y=a^{\left(t+\frac{1}{t}\right)} \end{gathered}$$

$$\begin{gathered} \Rightarrow \frac{dy}{dt}=\frac{d}{dt}\left[a^{\left(t+\frac{1}{t}\right)}\right] \\ \Rightarrow \frac{dy}{dt}=a^{\left(t+\frac{1}{t}\right)}\times \log a\frac{d}{dt}\left(t+\frac{t}{t}\right) \\ \Rightarrow \frac{dy}{dt}=a^{\left(t+\frac{1}{t}\right)}\times \log a\left(1-\frac{1}{t^2}\right)...\left(ii\right) \\ \mathrm{Dividing}\mathrm{equation}\left(\mathrm{ii}\right)\mathrm{by}\left(\mathrm{i}\right), \\ \frac{{\displaystyle \frac{dy}{dt}}}{{\displaystyle \frac{dx}{dt}}}=\frac{a^{\left(t+\frac{1}{t}\right)}\times \log a\left(1-\frac{1}{t^2}\right)}{a{\left(t+{\displaystyle \frac{1}{t}}\right)}^{a-1}\left(1-{\displaystyle \frac{1}{t^2}}\right)} \\ \Rightarrow \frac{dy}{dx}=\frac{a^{\left(t+\frac{1}{t}\right)}\times \log a}{a{\left(t+\frac{1}{t}\right)}^{a-1}} \end{gathered}$$