Differentiate. (a) $$ y=\frac{8}{\sin ^{-1}(x)} $$ (b) $$ f(x)=x \arctan (\sqrt{x}) $$ Solution $$ \begin{array}{l} \frac{d y}{d x}=\frac{d}{d x}\left[8\left(\sin ^{-1}(x) ight)^{-1} ight]=(\square) \frac{d}{d x}\left(\sin ^{-1}(x) ight. \ ext { (a) }=\square \ ext { (b) } \ ext { (b) } \ ext { (x) }=x(\square \arctan (\sqrt{x}) \ =\square\end{array} $$

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Differentiate. (a) $$ y=\frac{8}{\sin ^{-1}(x)} $$ (b) $$ f(x)=x \arctan (\sqrt{x}) $$ Solution $$ \begin{array}{l} \frac{d y}{d x}=\frac{d}{d x}\left[8\left(\sin ^{-1}(x)ight)^{-1}ight]=(\square) \frac{d}{d x}\left(\sin ^{-1}(x)ight. \ 	ext { (a) }=\square \ 	ext { (b) } \ 	ext { (b) } \ 	ext { (x) }=x(\square \arctan (\sqrt{x}) \ =\square\end{array} $$

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**(a) Differentiation of** $$ y=\frac{8}{\sin^{-1}(x)} $$

We write the function as
$$
y=8\left(\sin^{-1}(x)\right)^{-1}.
$$
Let 
$$
u=\sin^{-1}(x) \quad \text{so that} \quad y=8u^{-1}.
$$
Differentiate $$ y $$ with respect to $$ x $$ using the chain rule:
$$
\frac{dy}{dx}=8\frac{d}{dx}\left(u^{-1}\right)=-8u^{-2}\frac{du}{dx}.
$$
Recall that
$$
\frac{du}{dx}=\frac{d}{dx}\left(\sin^{-1}(x)\right)=\frac{1}{\sqrt{1-x^2}}.
$$
Thus,
$$
\frac{dy}{dx}=-8\left(\sin^{-1}(x)\right)^{-2}\frac{1}{\sqrt{1-x^2}} = -\frac{8}{\left(\sin^{-1}(x)\right)^2\sqrt{1-x^2}}.
$$

---

**(b) Differentiation of** $$ f(x)=x\arctan(\sqrt{x}) $$

We use the product rule with
$$
u = x \quad \text{and} \quad v = \arctan(\sqrt{x}).
$$
Then,
$$
f'(x) = u'v + uv'.
$$
Since
$$
u' = 1,
$$
we need to compute $$ v' $$. Let
$$
w=\sqrt{x} \quad \text{so that} \quad v=\arctan(w).
$$
Differentiate $$ v $$ with respect to $$ w $$ and then $$ w $$ with respect to $$ x $$:
$$
\frac{dv}{dw}=\frac{1}{1+w^2}=\frac{1}{1+x} \quad\text{(since $$w^2=x$$)},
$$
and
$$
\frac{dw}{dx}=\frac{1}{2\sqrt{x}}.
$$
Thus,
$$
v'=\frac{dv}{dw}\frac{dw}{dx}=\frac{1}{1+x}\cdot\frac{1}{2\sqrt{x}}=\frac{1}{2\sqrt{x}(1+x)}.
$$
Now apply the product rule:
$$
f'(x)=1\cdot\arctan(\sqrt{x})+x\cdot\frac{1}{2\sqrt{x}(1+x)}.
$$
Simplify the second term:
$$
\frac{x}{2\sqrt{x}(1+x)}=\frac{\sqrt{x}}{2(1+x)}.
$$
Thus, the derivative is
$$
f'(x)=\arctan(\sqrt{x})+\frac{\sqrt{x}}{2(1+x)}.
$$
**(a)** $$ \frac{dy}{dx} = -\frac{8}{\left(\sin^{-1}(x)\right)^2\sqrt{1-x^2}} $$ **(b)** $$ f'(x) = \arctan(\sqrt{x}) + \frac{\sqrt{x}}{2(1+x)} $$

Differentiate.
(a)
(b)
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