Mathematics - Differentiation Question with Solution | TestHub

$$\begin{array}{rl}& \text{ Let }x=\sqrt{a^{{\sin }^{-1}t}}\\ & \Rightarrow x^2=a^{{\sin }^{-1}t}\Rightarrow 2\log x={\sin }^{-1}t\cdot \log a\\ & \Rightarrow \frac{2}{x}=\frac{\log a}{\sqrt{1-t^2}}\cdot \frac{dt}{dx}\\ & \Rightarrow \frac{2\sqrt{1-t^2}}{x\log a}=\frac{dt}{dx}\end{array}$$ Now, let $y=\sqrt{a^{{\cos }^{-1}t}}$ $$\begin{array}{rl}& \Rightarrow 2\log y={\cos }^{-1}t\cdot \log a\\ & \Rightarrow \frac{2}{y}\cdot \frac{dy}{dx}=\frac{-\log a}{\sqrt{1-t^2}}\cdot \frac{dt}{dx}\end{array}$$ $\Rightarrow \frac{2}{y}\cdot \frac{dy}{dx}=\frac{-\log a}{\sqrt{1-t^2}}\times \frac{2\sqrt{1-t^2}}{x\log a}$ (from(1) $$\Rightarrow \frac{dy}{dx}=-\frac{y}{x}$$ Hence, $1+{\left(\frac{dy}{dx}\right)}^2=1+{\left(\frac{-y}{x}\right)}^2=\frac{x^2+y^2}{x^2}$