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

# Explanation:
## Step1: Differentiate $f(x)=\frac{x^{2}\sin(x)}{1 + x^{2}}$ using the quotient rule
The quotient rule is $(\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$, where $u = x^{2}\sin(x)$ and $v=1 + x^{2}$.
First, find $u^\prime$ using the product rule $(uv)^\prime=u^\prime v+uv^\prime$. Let $u_1=x^{2}$ and $u_2 = \sin(x)$, then $u^\prime=2x\sin(x)+x^{2}\cos(x)$ and $v^\prime = 2x$.
$$
\begin{align*}
f^\prime(x)&=\frac{(2x\sin(x)+x^{2}\cos(x))(1 + x^{2})-x^{2}\sin(x)\cdot2x}{(1 + x^{2})^{2}}\
&=\frac{2x\sin(x)+2x^{3}\sin(x)+x^{2}\cos(x)+x^{4}\cos(x)-2x^{3}\sin(x)}{(1 + x^{2})^{2}}\
&=\frac{2x\sin(x)+x^{2}\cos(x)+x^{4}\cos(x)}{(1 + x^{2})^{2}}
\end{align*}
$$
## Step2: Differentiate $f(x)=\sin^{2}(3x)\sin(4x^{5})$ using the product rule and chain rule
The product rule $(uv)^\prime=u^\prime v+uv^\prime$, where $u=\sin^{2}(3x)$ and $v = \sin(4x^{5})$.
For $u$, let $t=\sin(3x)$, then $u = t^{2}$, $u^\prime=2t\cdot3\cos(3x)=6\sin(3x)\cos(3x)$ by the chain rule.
For $v$, let $s = 4x^{5}$, then $v=\sin(s)$, $v^\prime=\cos(s)\cdot20x^{4}=20x^{4}\cos(4x^{5})$ by the chain rule.
$$
\begin{align*}
f^\prime(x)&=6\sin(3x)\cos(3x)\sin(4x^{5})+\sin^{2}(3x)\cdot20x^{4}\cos(4x^{5})\
&=3\sin(6x)\sin(4x^{5})+20x^{4}\sin^{2}(3x)\cos(4x^{5})
\end{align*}
$$
## Step3: Differentiate $f(x)=\sqrt{1+\sqrt{1+\sqrt{1 + x}}}$ using the chain rule
Let $y=\sqrt{1+\sqrt{1+\sqrt{1 + x}}}$, let $u_1=1+\sqrt{1+\sqrt{1 + x}}$, $u_2=1+\sqrt{1 + x}$, $u_3=1 + x$.
$y=\sqrt{u_1}$, $y^\prime=\frac{1}{2\sqrt{u_1}}\cdot u_1^\prime$.
$u_1^\prime=\frac{1}{2\sqrt{u_2}}\cdot u_2^\prime$, $u_2^\prime=\frac{1}{2\sqrt{u_3}}\cdot u_3^\prime$, and $u_3^\prime = 1$.
$$
\begin{align*}
f^\prime(x)&=\frac{1}{2\sqrt{1+\sqrt{1+\sqrt{1 + x}}}}\cdot\frac{1}{2\sqrt{1+\sqrt{1 + x}}}\cdot\frac{1}{2\sqrt{1 + x}}\
&=\frac{1}{8\sqrt{1 + x}\sqrt{1+\sqrt{1 + x}}\sqrt{1+\sqrt{1+\sqrt{1 + x}}}}
\end{align*}
$$

# Answer:
(a) $f^\prime(x)=\frac{2x\sin(x)+x^{2}\cos(x)+x^{4}\cos(x)}{(1 + x^{2})^{2}}$
(b) $f^\prime(x)=3\sin(6x)\sin(4x^{5})+20x^{4}\sin^{2}(3x)\cos(4x^{5})$
(c) $f^\prime(x)=\frac{1}{8\sqrt{1 + x}\sqrt{1+\sqrt{1 + x}}\sqrt{1+\sqrt{1+\sqrt{1 + x}}}}$

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QUESTION IMAGE

Question

Explanation:

Step1: Differentiate \(f(x)=\frac{x^{2}\sin(x)}{1 + x^{2}}\) using the quotient rule

Step2: Differentiate \(f(x)=\sin^{2}(3x)\sin(4x^{5})\) using the product rule and chain rule

Step3: Differentiate \(f(x)=\sqrt{1+\sqrt{1+\sqrt{1 + x}}}\) using the chain rule

Answer: