Question

differentiate the following functions. (a) 4pts. $f(x) = \frac{x^2 \sin(x)}{1 + x^2}$ (b) 4pts. $f(x) = \sin^2(3x) \sin(4x^5)$ (c) 4pts. $f(x) = \sqrt{1 + \sqrt{1 + \sqrt{1 + x}}}$

Explanation:

Part (a)

Step 1: Identify the rule

We use the quotient rule: if \( f(x)=\frac{u(x)}{v(x)} \), then \( f'(x)=\frac{u'(x)v(x)-u(x)v'(x)}{v(x)^2} \). Here, \( u(x) = x^2\sin(x) \) and \( v(x)=1 + x^2 \).

Step 2: Differentiate \( u(x) \)

Use the product rule: if \( u(x)=a(x)b(x) \), then \( u'(x)=a'(x)b(x)+a(x)b'(x) \). Let \( a(x)=x^2 \), \( b(x)=\sin(x) \). Then \( a'(x) = 2x \), \( b'(x)=\cos(x) \). So \( u'(x)=2x\sin(x)+x^2\cos(x) \).

Step 3: Differentiate \( v(x) \)

\( v'(x)=2x \) (since derivative of \( 1 \) is \( 0 \) and derivative of \( x^2 \) is \( 2x \)).

Step 4: Apply quotient rule

$$
LATEXBLOCK0
$$

Step 1: Identify the rule

Use the product rule: if \( f(x)=u(x)v(x) \), then \( f'(x)=u'(x)v(x)+u(x)v'(x) \). Here, \( u(x)=\sin^2(3x) \) and \( v(x)=\sin(4x^5) \).

Step 2: Differentiate \( u(x) \)

Use the chain rule: let \( u = \sin(3x) \), so \( u(x)=u^2 \). Then \( u'(x)=2u\cdot u' \). Now, \( u = \sin(3x) \), so derivative of \( u \) with respect to \( x \) is \( 3\cos(3x) \). Thus, \( u'(x)=2\sin(3x)\cdot 3\cos(3x)=6\sin(3x)\cos(3x) \).

Step 3: Differentiate \( v(x) \)

Use the chain rule: let \( t = 4x^5 \), so \( v(x)=\sin(t) \). Then \( v'(x)=\cos(t)\cdot t' \). Derivative of \( t = 4x^5 \) is \( 20x^4 \). So \( v'(x)=\cos(4x^5)\cdot 20x^4 = 20x^4\cos(4x^5) \).

Step 4: Apply product rule

$$
LATEXBLOCK0
$$

Step 1: Let \( y = f(x)=\sqrt{1+\sqrt{1+\sqrt{1 + x}}} \)

Let \( u = 1+\sqrt{1+\sqrt{1 + x}} \), so \( y=\sqrt{u}=u^{\frac{1}{2}} \).

Step 2: Differentiate \( y \) with respect to \( u \)

Using power rule: \( \frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}=\frac{1}{2\sqrt{u}}=\frac{1}{2\sqrt{1+\sqrt{1+\sqrt{1 + x}}}} \).

Step 3: Differentiate \( u \) with respect to \( x \)

\( u = 1 + v \), where \( v=\sqrt{1+\sqrt{1 + x}} \). So \( \frac{du}{dx}=\frac{dv}{dx} \). Now, for \( v=\sqrt{1+\sqrt{1 + x}} \), let \( w = 1+\sqrt{1 + x} \), so \( v=\sqrt{w}=w^{\frac{1}{2}} \). Then \( \frac{dv}{dw}=\frac{1}{2}w^{-\frac{1}{2}}=\frac{1}{2\sqrt{w}}=\frac{1}{2\sqrt{1+\sqrt{1 + x}}} \).

Step 4: Differentiate \( w \) with respect to \( x \)

\( w = 1 + z \), where \( z=\sqrt{1 + x} \). So \( \frac{dw}{dx}=\frac{dz}{dx} \). For \( z=\sqrt{1 + x}=(1 + x)^{\frac{1}{2}} \), using power rule, \( \frac{dz}{dx}=\frac{1}{2}(1 + x)^{-\frac{1}{2}}=\frac{1}{2\sqrt{1 + x}} \).

Step 5: Apply chain rule

First, \( \frac{dv}{dx}=\frac{dv}{dw}\cdot\frac{dw}{dx}=\frac{1}{2\sqrt{1+\sqrt{1 + x}}}\cdot\frac{1}{2\sqrt{1 + x}}=\frac{1}{4\sqrt{1+\sqrt{1 + x}}\sqrt{1 + x}} \)

Then, \( \frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}=\frac{1}{2\sqrt{1+\sqrt{1+\sqrt{1 + x}}}}\cdot\frac{1}{4\sqrt{1+\sqrt{1 + x}}\sqrt{1 + x}} \)

Simplify:

$$ f'(x)=\frac{1}{8\sqrt{1+\sqrt{1+\sqrt{1 + x}}}\sqrt{1+\sqrt{1 + x}}\sqrt{1 + x}} $$

Answer:

\( f'(x)=\frac{2x\sin(x)+x^2\cos(x)(1 + x^2)}{(1 + x^2)^2} \) (or simplified form as above)

Part (b)