Question

9.
$f(x) = \frac{x}{4} \sin^{-1}(x)$

10.
$x^2 y = (y + 2) + xy \sin(x)$

Explanation:

Assuming we need to find the derivative of \( f(x)=\frac{x}{4}\sin^{-1}(x) \) (for problem 9, as problem 10 is an implicit equation, but let's focus on 9 first). We use the product rule: if \( f(x) = u(x)v(x) \), then \( f'(x)=u'(x)v(x)+u(x)v'(x) \).

Step1: Identify \( u \) and \( v \)

Let \( u(x)=\frac{x}{4} \), so \( u'(x)=\frac{1}{4} \). Let \( v(x)=\sin^{-1}(x) \), and the derivative of \( \sin^{-1}(x) \) is \( \frac{1}{\sqrt{1 - x^2}} \), so \( v'(x)=\frac{1}{\sqrt{1 - x^2}} \).

Step2: Apply product rule

Using \( f'(x)=u'(x)v(x)+u(x)v'(x) \), substitute the values:
\( f'(x)=\frac{1}{4}\cdot\sin^{-1}(x)+\frac{x}{4}\cdot\frac{1}{\sqrt{1 - x^2}} \)
Simplify: \( f'(x)=\frac{\sin^{-1}(x)}{4}+\frac{x}{4\sqrt{1 - x^2}} \)

For problem 10 (\( x^2y=(y + 2)+xy\sin(x) \)), we use implicit differentiation.

Step1: Differentiate both sides w.r. to \( x \)

Left side: Use product rule on \( x^2y \), derivative is \( 2xy + x^2y' \).

Right side: Derivative of \( (y + 2) \) is \( y' \), derivative of \( xy\sin(x) \): use product rule (three factors, but treat as \( (xy)\sin(x) \)). Let \( u = xy \), \( v=\sin(x) \). Then \( u'=y + xy' \), \( v'=\cos(x) \). So derivative of \( xy\sin(x) \) is \( (y + xy')\sin(x)+xy\cos(x) \). So right side derivative: \( y' + (y + xy')\sin(x)+xy\cos(x) \).

Step2: Collect \( y' \) terms

\( 2xy + x^2y' = y' + (y + xy')\sin(x)+xy\cos(x) \)

Expand right side: \( y' + y\sin(x)+xy'\sin(x)+xy\cos(x) \)

Bring all \( y' \) terms to left and others to right:

\( x^2y' - y' - xy'\sin(x)=y\sin(x)+xy\cos(x)-2xy \)

Factor \( y' \):

\( y'(x^2 - 1 - xy\sin(x))=y\sin(x)+xy\cos(x)-2xy \)

Then \( y'=\frac{y\sin(x)+xy\cos(x)-2xy}{x^2 - 1 - xy\sin(x)} \) (we can factor numerator and denominator if needed, but this is the derivative).

Answer:

For \( f(x)=\frac{x}{4}\sin^{-1}(x) \), derivative is \( \boldsymbol{\frac{\sin^{-1}(x)}{4}+\frac{x}{4\sqrt{1 - x^2}}} \).

For \( x^2y=(y + 2)+xy\sin(x) \), derivative \( y' \) is \( \boldsymbol{\frac{y\sin(x)+xy\cos(x)-2xy}{x^2 - 1 - xy\sin(x)}} \) (or simplified form).