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 \( f^\prime(x)=\frac{u^\prime(x)v(x)-u(x)v^\prime(x)}{v(x)^2} \). Here, \( u(x) = x^2\sin(x) \) and \( v(x)=1 + x^2 \).

Step2: Differentiate \( u(x) \) (product rule)

Product rule: \( (ab)^\prime=a^\prime b+ab^\prime \). For \( u(x)=x^2\sin(x) \), \( a = x^2 \), \( b=\sin(x) \). So \( u^\prime(x)=2x\sin(x)+x^2\cos(x) \).

Step3: Differentiate \( v(x) \)

\( v(x)=1 + x^2 \), so \( v^\prime(x)=2x \).

Step4: Apply quotient rule

\[

$$\begin{align*} f^\prime(x)&=\frac{(2x\sin(x)+x^2\cos(x))(1 + x^2)-x^2\sin(x)(2x)}{(1 + x^2)^2}\\ &=\frac{2x\sin(x)(1 + x^2)+x^2\cos(x)(1 + x^2)-2x^3\sin(x)}{(1 + x^2)^2}\\ &=\frac{2x\sin(x)+2x^3\sin(x)+x^2\cos(x)+x^4\cos(x)-2x^3\sin(x)}{(1 + x^2)^2}\\ &=\frac{2x\sin(x)+x^2\cos(x)+x^4\cos(x)}{(1 + x^2)^2}\\ &=\frac{x(2\sin(x)+x\cos(x)+x^3\cos(x))}{(1 + x^2)^2} \end{align*}$$

\]

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: \( f^\prime(x)=u^\prime(x)v(x)+u(x)v^\prime(x) \). Then use chain rule for \( u^\prime \) and \( v^\prime \).

Step2: Differentiate \( u(x) \) (chain rule)

Let \( u=\sin(3x) \), so \( u(x)=u^2 \). Then \( u^\prime(x)=2u\cdot u^\prime=2\sin(3x)\cdot 3\cos(3x)=6\sin(3x)\cos(3x) \).

Step3: Differentiate \( v(x) \) (chain rule)

Let \( v=\sin(w) \), \( w = 4x^5 \). Then \( v^\prime(x)=\cos(w)\cdot w^\prime=\cos(4x^5)\cdot 20x^4=20x^4\cos(4x^5) \).

Step4: Apply product rule

\[

$$\begin{align*} f^\prime(x)&=6\sin(3x)\cos(3x)\sin(4x^5)+\sin^2(3x)\cdot 20x^4\cos(4x^5)\\ &=6\sin(3x)\cos(3x)\sin(4x^5)+20x^4\sin^2(3x)\cos(4x^5) \end{align*}$$

\]

Step1: Let \( y = f(x)=\sqrt{1+\sqrt{1+\sqrt{1 + x}}} \). Use chain rule multiple times. Let \( u = 1+\sqrt{1+\sqrt{1 + x}} \), so \( y=\sqrt{u}=u^{\frac{1}{2}} \). Then \( y^\prime=\frac{1}{2}u^{-\frac{1}{2}}\cdot u^\prime \).

Step2: Differentiate \( u \). Let \( v = 1+\sqrt{1 + x} \), so \( u = 1+\sqrt{v}=1 + v^{\frac{1}{2}} \). Then \( u^\prime=\frac{1}{2}v^{-\frac{1}{2}}\cdot v^\prime \).

Step3: Differentiate \( v \). Let \( w = 1 + x \), so \( v = 1+\sqrt{w}=1 + w^{\frac{1}{2}} \). Then \( v^\prime=\frac{1}{2}w^{-\frac{1}{2}}\cdot w^\prime \).

Step4: Differentiate \( w \)

\( w = 1 + x \), so \( w^\prime=1 \).

Step5: Back - substitute

First, \( v^\prime=\frac{1}{2}(1 + x)^{-\frac{1}{2}}\cdot1=\frac{1}{2\sqrt{1 + x}} \).

Then, \( u^\prime=\frac{1}{2}(1+\sqrt{1 + x})^{-\frac{1}{2}}\cdot\frac{1}{2\sqrt{1 + x}}=\frac{1}{4\sqrt{1+\sqrt{1 + x}}\sqrt{1 + x}} \).

Then, \( y^\prime=\frac{1}{2}(1+\sqrt{1+\sqrt{1 + x}})^{-\frac{1}{2}}\cdot\frac{1}{4\sqrt{1+\sqrt{1 + x}}\sqrt{1 + x}} \)

Simplify:
\[
f^\prime(x)=\frac{1}{8\sqrt{1+\sqrt{1+\sqrt{1 + x}}}\sqrt{1+\sqrt{1 + x}}\sqrt{1 + x}}
\]

Answer:

\( \frac{2x\sin(x)+x^2\cos(x)+x^4\cos(x)}{(1 + x^2)^2} \) (or factored form as above)

Part (b)