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. $f(x) = \sqrt{1 + \sqrt{1 + \sqrt{1 + x}}}$

Question

Accepted Answer

### Part (a)
# Explanation:
## Step1: Identify the 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) $
Use the product rule: if $ u(x)=a(x)b(x) $, then $ u^\prime(x)=a^\prime(x)b(x)+a(x)b^\prime(x) $. Let $ a(x)=x^{2} $, $ b(x)=\sin(x) $. Then $ a^\prime(x) = 2x $, $ b^\prime(x)=\cos(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
$$
\begin{align*}
f^\prime(x)&=\frac{(2x\sin(x)+x^{2}\cos(x))(1 + x^{2})-x^{2}\sin(x)(2x)}{(1 + x^{2})^{2}}\
&=\frac{2x\sin(x)(1 + x^{2})+x^{2}\cos(x)(1 + x^{2})-2x^{3}\sin(x)}{(1 + x^{2})^{2}}\
&=\frac{2x\sin(x)+2x^{3}\sin(x)+x^{2}\cos(x)+x^{4}\cos(x)-2x^{3}\sin(x)}{(1 + x^{2})^{2}}\
&=\frac{2x\sin(x)+x^{2}\cos(x)+x^{4}\cos(x)}{(1 + x^{2})^{2}}\
&=\frac{x(2\sin(x)+x\cos(x)+x^{3}\cos(x))}{(1 + x^{2})^{2}}
\end{align*}
$$

# Answer:
$ f^\prime(x)=\frac{2x\sin(x)+x^{2}\cos(x)+x^{4}\cos(x)}{(1 + x^{2})^{2}} $ (or simplified form)

### Part (b)
# Explanation:
## Step1: Identify the rule
We use the product rule: if $ f(x)=u(x)v(x) $, then $ f^\prime(x)=u^\prime(x)v(x)+u(x)v^\prime(x) $. Here, $ u(x)=\sin^{2}(3x) $ and $ v(x)=\sin(4x^{5}) $.

## Step2: Differentiate $ u(x) $
Use the chain rule: if $ u(x)=[g(x)]^{n} $, then $ u^\prime(x)=n[g(x)]^{n - 1}g^\prime(x) $. Let $ g(x)=\sin(3x) $, $ n = 2 $. First, $ g^\prime(x)=3\cos(3x) $ (by chain rule: derivative of $ \sin(3x) $ is $ \cos(3x)\times3 $). So $ u^\prime(x)=2\sin(3x)\times3\cos(3x)=6\sin(3x)\cos(3x) $.

## Step3: Differentiate $ v(x) $
Use the chain rule: if $ v(x)=\sin(h(x)) $, then $ v^\prime(x)=\cos(h(x))h^\prime(x) $. Let $ h(x)=4x^{5} $, so $ h^\prime(x)=20x^{4} $. Then $ v^\prime(x)=\cos(4x^{5})\times20x^{4}=20x^{4}\cos(4x^{5}) $.

## Step4: Apply product rule
$$
\begin{align*}
f^\prime(x)&=6\sin(3x)\cos(3x)\times\sin(4x^{5})+\sin^{2}(3x)\times20x^{4}\cos(4x^{5})\
&=3\sin(6x)\sin(4x^{5})+20x^{4}\sin^{2}(3x)\cos(4x^{5})
\end{align*}
$$
(using the double - angle formula $ \sin(2\theta)=2\sin\theta\cos\theta $, so $ 6\sin(3x)\cos(3x)=3\sin(6x) $)

# Answer:
$ f^\prime(x)=3\sin(6x)\sin(4x^{5})+20x^{4}\sin^{2}(3x)\cos(4x^{5}) $

### Part (c)
# Explanation:
## Step1: Let $ y = f(x)=\sqrt{1+\sqrt{1+\sqrt{1 + x}}} $
We use the chain rule multiple times. Let $ y=\sqrt{u} $, where $ u = 1+\sqrt{v} $, $ v = 1+\sqrt{w} $, $ w=1 + x $.

## Step2: Differentiate $ y $ with respect to $ u $
$ y=\sqrt{u}=u^{\frac{1}{2}} $, so $ \frac{dy}{du}=\frac{1}{2\sqrt{u}} $.

## Step3: Differentiate $ u $ with respect to $ v $
$ u = 1+\sqrt{v}=1 + v^{\frac{1}{2}} $, so $ \frac{du}{dv}=\frac{1}{2\sqrt{v}} $.

## Step4: Differentiate $ v $ with respect to $ w $
$ v = 1+\sqrt{w}=1 + w^{\frac{1}{2}} $, so $ \frac{dv}{dw}=\frac{1}{2\sqrt{w}} $.

## Step5: Differentiate $ w $ with respect to $ x $
$ w=1 + x $, so $ \frac{dw}{dx}=1 $.

## Step6: Apply chain rule $ \frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dv}\times\frac{dv}{dw}\times\frac{dw}{dx} $
Substitute back $ u = 1+\sqrt{v} $, $ v = 1+\sqrt{w} $, $ w = 1 + x $:

$$
\begin{align*}
\frac{dy}{dx}&=\frac{1}{2\sqrt{1+\sqrt{1+\sqrt{1 + x}}}}\times\frac{1}{2\sqrt{1+\sqrt{1 + x}}}\times\frac{1}{2\sqrt{1 + x}}\times1\
&=\frac{1}{8\sqrt{1+\sqrt{1+\sqrt{1 + x}}}\sqrt{1+\sqrt{1 + x}}\sqrt{1 + x}}
\end{align*}
$$

# Answer:
$ f^\prime(x)=\frac{1}{8\sqrt{1+\sqrt{1+\sqrt{1 + x}}}\sqrt{1+\sqrt{1 + x}}\sqrt{1 + x}} $

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QUESTION IMAGE

Question

Explanation:

Part (a)

Step1: Identify the rule

Step2: Differentiate \( u(x) \)

Step3: Differentiate \( v(x) \)

Step4: Apply quotient rule

Step1: Identify the rule

Step2: Differentiate \( u(x) \)

Step3: Differentiate \( v(x) \)

Step4: Apply product rule

Step1: Let \( y = f(x)=\sqrt{1+\sqrt{1+\sqrt{1 + x}}} \)

Step2: Differentiate \( y \) with respect to \( u \)

Step3: Differentiate \( u \) with respect to \( v \)

Step4: Differentiate \( v \) with respect to \( w \)

Step5: Differentiate \( w \) with respect to \( x \)

Step6: Apply chain rule \( \frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dv}\times\frac{dv}{dw}\times\frac{dw}{dx} \)

Answer:

Part (b)