$$ \left. \begin{array} { l } { h ( x ) = \sqrt { 1 - x ^ { 2 } } \sin ^ { - 1 } ( x ) } \ { h ^ { \prime } ( x ) = \square } \end{array} ight. $$

Question

$$ \left. \begin{array} { l } { h ( x ) = \sqrt { 1 - x ^ { 2 } } \sin ^ { - 1 } ( x ) } \ { h ^ { \prime } ( x ) = \square } \end{array} ight. $$

UpStudy AI · Accepted Answer

We have
$$
h(x)=\sqrt{1-x^2}\,\sin^{-1}(x).
$$
We will use the product rule for differentiation:
$$
h'(x)= f'(x)g(x)+ f(x)g'(x),
$$
where
$$
f(x)=\sqrt{1-x^2} \quad \text{and} \quad g(x)=\sin^{-1}(x).
$$

1. Differentiate $$ f(x)= (1-x^2)^{\frac{1}{2}} $$ using the chain rule:
   $$
   f'(x)= \frac{1}{2}(1-x^2)^{-\frac{1}{2}} \cdot (-2x)= -\frac{x}{\sqrt{1-x^2}}.
   $$
   
2. Differentiate $$ g(x)=\sin^{-1}(x) $$:
   $$
   g'(x)=\frac{1}{\sqrt{1-x^2}}.
   $$

3. Apply the product rule:
   $$
   h'(x)= \left(-\frac{x}{\sqrt{1-x^2}}\right)\sin^{-1}(x) + \sqrt{1-x^2}\cdot\frac{1}{\sqrt{1-x^2}}.
   $$
   Simplify the expression:
   $$
   h'(x)= -\frac{x\sin^{-1}(x)}{\sqrt{1-x^2}} + 1.
   $$

Thus, the derivative is
$$
h'(x)=1-\frac{x\sin^{-1}(x)}{\sqrt{1-x^2}}.
$$
The derivative of $$ h(x) = \sqrt{1 - x^2} \sin^{-1}(x) $$ is: $$ h'(x) = 1 - \frac{x \sin^{-1}(x)}{\sqrt{1 - x^2}}. $$

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