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 quotient rule
The quotient rule states that if $y=\frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u = x^{2}\sin(x)$, $u^\prime=2x\sin(x)+x^{2}\cos(x)$ (using product rule $(uv)^\prime = u^\prime v+uv^\prime$ with $u = x^{2}$, $u^\prime=2x$ and $v=\sin(x)$, $v^\prime=\cos(x)$), and $v = 1 + x^{2}$, $v^\prime = 2x$.
$$
\begin{align*}
f^\prime(x)&=\frac{(2x\sin(x)+x^{2}\cos(x))(1 + x^{2})-x^{2}\sin(x)\cdot2x}{(1 + x^{2})^{2}}\
&=\frac{2x\sin(x)+2x^{3}\sin(x)+x^{2}\cos(x)+x^{4}\cos(x)-2x^{3}\sin(x)}{(1 + x^{2})^{2}}\
&=\frac{2x\sin(x)+x^{2}\cos(x)+x^{4}\cos(x)}{(1 + x^{2})^{2}}
\end{align*}
$$
## Step2: Differentiate $f(x)=\sin^{2}(3x)\sin(4x^{5})$ using product rule and chain rule
The product rule: if $y = u\cdot v$, then $y^\prime=u^\prime v+uv^\prime$. Let $u=\sin^{2}(3x)$, using chain rule $u^\prime = 2\sin(3x)\cos(3x)\cdot3=3\sin(6x)$ (since $\sin(2\alpha)=2\sin\alpha\cos\alpha$). Let $v=\sin(4x^{5})$, using chain rule $v^\prime=\cos(4x^{5})\cdot20x^{4}$.
$$
\begin{align*}
f^\prime(x)&=3\sin(6x)\sin(4x^{5})+20x^{4}\cos(4x^{5})\sin^{2}(3x)
\end{align*}
$$
## Step3: Differentiate $f(x)=\sqrt{1+\sqrt{1+\sqrt{1 + x}}}$ using chain rule
Let $y=\sqrt{u}$, $u = 1+\sqrt{v}$, $v=1+\sqrt{w}$, $w = 1 + x$. Then $\frac{dy}{du}=\frac{1}{2\sqrt{u}}$, $\frac{du}{dv}=\frac{1}{2\sqrt{v}}$, $\frac{dv}{dw}=\frac{1}{2\sqrt{w}}$, $\frac{dw}{dx}=1$.
$$
\begin{align*}
f^\prime(x)&=\frac{1}{2\sqrt{1+\sqrt{1+\sqrt{1 + x}}}}\cdot\frac{1}{2\sqrt{1+\sqrt{1 + x}}}\cdot\frac{1}{2\sqrt{1 + x}}\
&=\frac{1}{8\sqrt{(1 + x)(1+\sqrt{1 + x})(1+\sqrt{1+\sqrt{1 + x}})}}
\end{align*}
$$

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

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QUESTION IMAGE

Question

Explanation:

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

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

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

Answer: