If $x = a(\log t)$ and $f(t) = a(\sin^{-1} t)$ where $a$ is a constant, find $\frac{df(t)}{dx}$.

$\frac{t}{\sqrt{1-t^2}}$

The correct answer is Option (1) → $\frac{t}{\sqrt{1-t^2}}$ ##

Given:

$f(t) = a \sin^{-1}(t)$

The derivative of $f(t)$ with respect to $t$ is:

$\frac{df}{dt} = a \cdot \frac{d}{dt}(\sin^{-1}(t)) = \frac{a}{\sqrt{1 - t^2}}$

Given:

$x = a \log t$

The derivative of $x$ with respect to $t$ is:

$\frac{dx}{dt} = a \cdot \frac{1}{t} = \frac{a}{t}$

So, $\frac{df}{dx} = \frac{df}{dt} \cdot \frac{dt}{dx} = \frac{a}{\sqrt{1 - t^2}} \cdot \frac{t}{a}$

$\frac{df}{dx} = \frac{t}{\sqrt{1 - t^2}}$