Find the derivative of the function $$ y=\sin ^{-1}(\sqrt{x}) $$ a) $$ y^{\prime}=\frac{-1}{2 x \sqrt{1-x^{2}}} $$ b) $$ y^{\prime}=\frac{1}{\sqrt{1-x^{2}}} $$ c) $$ y^{\prime}=\frac{1}{2 \sqrt{x} \sqrt{1-x}} $$ d) $$ y^{\prime}=\frac{-1}{\sqrt{x(1-x)^{2}}} $$

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UpStudy AI · Accepted Answer

We are given
$$
y = \sin^{-1}(\sqrt{x}).
$$
Let
$$
u = \sqrt{x}.
$$
Then
$$
y = \sin^{-1}(u).
$$

Step 1. Differentiate using the chain rule. We know that
$$
\frac{d}{du}(\sin^{-1}(u)) = \frac{1}{\sqrt{1-u^2}},
$$
and
$$
\frac{du}{dx} = \frac{d}{dx}(\sqrt{x}) = \frac{1}{2\sqrt{x}}.
$$

Step 2. Apply the chain rule:
$$
y' = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{1}{\sqrt{1-u^2}} \cdot \frac{1}{2\sqrt{x}}.
$$

Step 3. Substitute back $$ u = \sqrt{x} $$:
$$
y' = \frac{1}{2\sqrt{x} \sqrt{1-(\sqrt{x})^2}}.
$$
Since $$(\sqrt{x})^2 = x$$, we have
$$
y' = \frac{1}{2\sqrt{x} \sqrt{1-x}}.
$$

Thus, the derivative is:
$$
y' = \frac{1}{2\sqrt{x}\sqrt{1-x}},
$$
which corresponds to option c).
The derivative of $$ y = \sin^{-1}(\sqrt{x}) $$ is $$ y' = \frac{1}{2\sqrt{x}\sqrt{1-x}} $$.

Find the derivative of the function

a)
b)
c)
d)

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