Question

Differentiate \(y = e^{\cos^{-1}\sqrt{1 - x^2}}\) with respect to \(x\) and hence show that \(\frac{dy}{dx} = \frac{y}{\sqrt{1 - x^2}}\).

Hint:

Rewrite \(\cos^{-1}\sqrt{1 - x^2}\) as \(\sin^{-1}x\) (for \(0 < x < 1\)), or take logs and use the chain rule.

Answer 1

Step 1: Take the logarithm:
We have \(y = e^{\cos^{-1}\sqrt{1 - x^2}}\).
Taking natural log on both sides gives \(\log y = \cos^{-1}\sqrt{1 - x^2}\).

Step 2: Differentiate both sides:
Let \(u = \sqrt{1 - x^2}\). Then \(\frac{du}{dx} = \frac{-x}{\sqrt{1 - x^2}}\).
Use \(\frac{d}{dx}\cos^{-1}u = \frac{-1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx}\).
\[ \frac{1}{y}\frac{dy}{dx} = \frac{-1}{\sqrt{1 - (1 - x^2)}} \cdot \frac{-x}{\sqrt{1 - x^2}} \]

Step 3: Simplify:
Here \(1 - u^2 = x^2\), so \(\sqrt{1 - u^2} = x\) for \(x > 0\) (the principal-value case taken in this question).
\[ \frac{1}{y}\frac{dy}{dx} = \frac{x}{x\sqrt{1 - x^2}} = \frac{1}{\sqrt{1 - x^2}} \]

Step 4: Multiply by y:
\[ \frac{dy}{dx} = \frac{y}{\sqrt{1 - x^2}} \]

Final Answer:
\[ \boxed{\frac{dy}{dx} = \frac{y}{\sqrt{1 - x^2}}} \] This is what we had to show.