Question:

Given the function $y = e^{\sin^{-1}x} + e^{\cos^{-1}x}$. Step 1: Differentiate $y$ with respect to $x$. To find $\dfrac{dy}{dx}$, we differentiate each term using the chain rule. Recall that $\dfrac{d}{du}(e^u) = e^u$ and $\dfrac{d}{dx}(\sin^{-1}x) = \dfrac{1}{\sqrt{1-x^2}}$, and $\dfrac{d}{dx}(\cos^{-1}x) = \dfrac{-1}{\sqrt{1-x^2}}$. $$ \dfrac{dy}{dx} = \dfrac{d}{dx}(e^{\sin^{-1}x}) + \dfrac{d}{dx}(e^{\cos^{-1}x}) $$ Applying the chain rule: $$ \dfrac{d}{dx}(e^{\sin^{-1}x}) = e^{\sin^{-1}x} \cdot \dfrac{d}{dx}(\sin^{-1}x) = e^{\sin^{-1}x} \cdot \dfrac{1}{\sqrt{1-x^2}} $$ $$ \dfrac{d}{dx}(e^{\cos^{-1}x}) = e^{\cos^{-1}x} \cdot \dfrac{d}{dx}(\cos^{-1}x) = e^{\cos^{-1}x} \cdot \dfrac{-1}{\sqrt{1-x^2}} $$ Combining these results, we get: $$ \dfrac{dy}{dx} = \dfrac{e^{\sin^{-1}x}}{\sqrt{1-x^2}} - \dfrac{e^{\cos^{-1}x}}{\sqrt{1-x^2}} $$ $$ \dfrac{dy}{dx} = \dfrac{e^{\sin^{-1}x} - e^{\cos^{-1}x}}{\sqrt{1-x^2}} $$ Step 2: Analyze the domain of existence for $\dfrac{dy}{dx}$. The expression for $\dfrac{dy}{dx}$ contains the term $\dfrac{1}{\sqrt{1-x^2}}$. For this term to be defined, the expression under the square root must be positive, i.e., $1-x^2 > 0$. This implies $x^2 < 1$, which means $-1 < x < 1$. Therefore, $\dfrac{dy}{dx}$ exists for all $x \in (-1, 1)$. At $x = \pm 1$, the denominator $\sqrt{1-x^2}$ becomes zero, causing the derivative to be undefined (or to approach infinity). Step 3: Evaluate the properties of the derivative. Based on the derivation: 1. The derivative is given by the formula: $$ \dfrac{dy}{dx} = \dfrac{e^{\sin^{-1}x} - e^{\cos^{-1}x}}{\sqrt{1-x^2}} $$ This formula is valid for $x \in (-1, 1)$. 2. Rearranging the formula from Step 1, we can multiply both sides by $\sqrt{1-x^2}$: $$ \sqrt{1-x^2} \dfrac{dy}{dx} = e^{\sin^{-1}x} - e^{\cos^{-1}x} $$ This identity is also valid for $x \in (-1, 1)$. 3. The derivative $\dfrac{dy}{dx}$ exists for $x \in (-1, 1)$. It does not exist at the endpoints $x = -1$ and $x = 1$, as the one-sided limits of the derivative approach infinity at these points. Therefore, the statement that $\dfrac{dy}{dx}$ exists for all $x \in [-1, 1]$ is false.