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:

Step1: Recall quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. For $f(x)=\frac{x^{2}\sin(x)}{1 + x^{2}}$, let $u = x^{2}\sin(x)$ and $v=1 + x^{2}$. First, find $u^\prime$ using the product - rule $(uv)^\prime=u^\prime v+uv^\prime$. If $u = x^{2}$ and $v=\sin(x)$, then $u^\prime = 2x\sin(x)+x^{2}\cos(x)$ and $v^\prime = 2x$.
$f^\prime(x)=\frac{(2x\sin(x)+x^{2}\cos(x))(1 + x^{2})-x^{2}\sin(x)\cdot2x}{(1 + x^{2})^{2}}=\frac{2x\sin(x)+2x^{3}\sin(x)+x^{2}\cos(x)+x^{4}\cos(x)-2x^{3}\sin(x)}{(1 + x^{2})^{2}}=\frac{2x\sin(x)+x^{2}\cos(x)+x^{4}\cos(x)}{(1 + x^{2})^{2}}$

Step2: Recall product - rule and chain - rule

For $f(x)=\sin^{2}(3x)\sin(4x^{5})$, let $u=\sin^{2}(3x)$ and $v = \sin(4x^{5})$. First, find $u^\prime$ using the chain - rule. If $y = u^{2}$ and $u=\sin(3x)$, then $\frac{dy}{du}=2u$ and $\frac{du}{dx}=3\cos(3x)$, so $u^\prime=2\sin(3x)\cdot3\cos(3x)=6\sin(3x)\cos(3x)$. And for $v=\sin(4x^{5})$, using the chain - rule, if $y=\sin(u)$ and $u = 4x^{5}$, then $v^\prime=\cos(4x^{5})\cdot20x^{4}$.
By the product - rule $(uv)^\prime=u^\prime v+uv^\prime$, we have $f^\prime(x)=6\sin(3x)\cos(3x)\sin(4x^{5})+20x^{4}\sin^{2}(3x)\cos(4x^{5})$

Step3: Use chain - rule repeatedly

Let $y = f(x)=\sqrt{1+\sqrt{1+\sqrt{1 + x}}}$. Let $u = 1+\sqrt{1+\sqrt{1 + x}}$, so $y=\sqrt{u}=u^{\frac{1}{2}}$, then $\frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}$.
Let $v = 1+\sqrt{1 + x}$, then $u = 1 + \sqrt{v}$, and $\frac{du}{dv}=\frac{1}{2}v^{-\frac{1}{2}}$. Let $w=1 + x$, then $v = 1+\sqrt{w}$, and $\frac{dv}{dw}=\frac{1}{2}w^{-\frac{1}{2}}$, and $\frac{dw}{dx}=1$.
By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dv}\cdot\frac{dv}{dw}\cdot\frac{dw}{dx}=\frac{1}{2}(1+\sqrt{1+\sqrt{1 + x}})^{-\frac{1}{2}}\cdot\frac{1}{2}(1+\sqrt{1 + x})^{-\frac{1}{2}}\cdot\frac{1}{2}(1 + x)^{-\frac{1}{2}}\cdot1=\frac{1}{8\sqrt{1+\sqrt{1+\sqrt{1 + x}}}\sqrt{1+\sqrt{1 + x}}\sqrt{1 + x}}$

Answer:

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