Question

problem 1.
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)

Step1: Identify Quotient 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 \( v(x)=1 + x^{2} \).

Step2: Differentiate \( u(x) \) (Product Rule)

Product rule: \( (uv)'=u'v + uv' \). For \( u(x)=x^{2}\sin(x) \), let \( u_1 = x^{2} \), \( u_2=\sin(x) \). Then \( u_1' = 2x \), \( u_2'=\cos(x) \). So \( u'(x)=2x\sin(x)+x^{2}\cos(x) \).

Step3: Differentiate \( v(x) \)

\( v(x)=1 + x^{2} \), so \( v'(x)=2x \).

Step4: Apply Quotient Rule

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Step1: Identify Product and Chain Rules

We have \( f(x)=u(x)v(x) \) where \( u(x)=\sin^{2}(3x) \) and \( v(x)=\sin(4x^{5}) \). First, use product rule: \( f'(x)=u'(x)v(x)+u(x)v'(x) \). Then use chain rule for \( u(x) \) and \( v(x) \).

Step2: Differentiate \( u(x) \) (Chain Rule)

Let \( u(x)=(\sin(3x))^{2} \). Let \( y = \sin(3x) \), so \( u = y^{2} \). Then \( \frac{du}{dy}=2y \), \( \frac{dy}{dx}=3\cos(3x) \). So \( u'(x)=2\sin(3x)\cdot3\cos(3x)=6\sin(3x)\cos(3x) \) (or \( 3\sin(6x) \) using double - angle formula).

Step3: Differentiate \( v(x) \) (Chain Rule)

Let \( v(x)=\sin(4x^{5}) \). Let \( z = 4x^{5} \), so \( v=\sin(z) \). Then \( \frac{dv}{dz}=\cos(z) \), \( \frac{dz}{dx}=20x^{4} \). So \( v'(x)=\cos(4x^{5})\cdot20x^{4}=20x^{4}\cos(4x^{5}) \).

Step4: Apply Product Rule

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(We can also write \( 6\sin(3x)\cos(3x) \) as \( 3\sin(6x) \), so \( f'(x)=3\sin(6x)\sin(4x^{5})+20x^{4}\sin^{2}(3x)\cos(4x^{5}) \))

Step1: Use Chain Rule (Multiple Layers)

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

Step2: Differentiate \( u \)

Let \( u = 1 + v \), where \( v=\sqrt{1+\sqrt{1 + x}}=(1+(1 + x)^{\frac{1}{2}})^{\frac{1}{2}} \). Then \( u'=v' \). Now, for \( v \), let \( w = 1+\sqrt{1 + x}=(1+(1 + x)^{\frac{1}{2}}) \), so \( v = w^{\frac{1}{2}} \), \( v'=\frac{1}{2}w^{-\frac{1}{2}}\cdot w' \).

Step3: Differentiate \( w \)

Let \( w = 1+z \), where \( z=\sqrt{1 + x}=(1 + x)^{\frac{1}{2}} \). Then \( w'=z' \). For \( z=(1 + x)^{\frac{1}{2}} \), \( z'=\frac{1}{2}(1 + x)^{-\frac{1}{2}}\cdot1=\frac{1}{2\sqrt{1 + x}} \).

Step4: Back - substitute

• \( w = 1+(1 + x)^{\frac{1}{2}} \), so \( w'=\frac{1}{2\sqrt{1 + x}} \), then \( v'=\frac{1}{2}(1+(1 + x)^{\frac{1}{2}})^{-\frac{1}{2}}\cdot\frac{1}{2\sqrt{1 + x}}=\frac{1}{4\sqrt{1 + x}\sqrt{1+\sqrt{1 + x}}} \)
• \( u = 1+\sqrt{1+\sqrt{1 + x}} \), so \( u'=\frac{1}{4\sqrt{1 + x}\sqrt{1+\sqrt{1 + x}}} \)
• \( y = u^{\frac{1}{2}} \), so \( y'=\frac{1}{2}(1+\sqrt{1+\sqrt{1 + x}})^{-\frac{1}{2}}\cdot\frac{1}{4\sqrt{1 + x}\sqrt{1+\sqrt{1 + x}}} \)

Simplify the expression:

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

Answer:

\( f'(x)=\frac{2x\sin(x)+x^{2}\cos(x)+x^{4}\cos(x)}{(1 + x^{2})^{2}} \) (or simplified form)

Part (b)