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: Apply quotient - rule for (a)

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 $(uv)^\prime = u^\prime v+uv^\prime$ where $u = x^{2}$ and $v=\sin(x)$), and $v = 1 + x^{2}$, $v^\prime=2x$. So, $f^\prime(x)=\frac{(2x\sin(x)+x^{2}\cos(x))(1 + x^{2})-x^{2}\sin(x)\cdot2x}{(1 + x^{2})^{2}}=\frac{2x\sin(x)+2x^{3}\sin(x)+x^{2}\cos(x)+x^{4}\cos(x)-2x^{3}\sin(x)}{(1 + x^{2})^{2}}=\frac{2x\sin(x)+x^{2}\cos(x)+x^{4}\cos(x)}{(1 + x^{2})^{2}}$.

Step2: Apply product - rule and chain - rule for (b)

The product rule: if $y = uv$, then $y^\prime=u^\prime v+uv^\prime$. Let $u=\sin^{2}(3x)$ and $v = \sin(4x^{5})$. First, find $u^\prime$: using the chain - rule, if $y = u^{2}$ with $u=\sin(3x)$, 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 $v^\prime=\cos(4x^{5})\cdot20x^{4}$. Then $f^\prime(x)=6\sin(3x)\cos(3x)\sin(4x^{5})+20x^{4}\sin^{2}(3x)\cos(4x^{5})$.

Step3: Apply chain - rule for (c)

Let $y = f(x)=\sqrt{1 + \sqrt{1+\sqrt{1 + x}}}$, let $u = 1+\sqrt{1+\sqrt{1 + x}}$, then $y=\sqrt{u}$, $\frac{dy}{du}=\frac{1}{2\sqrt{u}}$. Let $v = 1+\sqrt{1 + x}$, then $u = 1+\sqrt{v}$, $\frac{du}{dv}=\frac{1}{2\sqrt{v}}$. Let $w = 1 + x$, then $v = 1+\sqrt{w}$, $\frac{dv}{dw}=\frac{1}{2\sqrt{w}}$, and $\frac{dw}{dx}=1$. So $f^\prime(x)=\frac{1}{2\sqrt{1+\sqrt{1+\sqrt{1 + x}}}}\cdot\frac{1}{2\sqrt{1+\sqrt{1 + x}}}\cdot\frac{1}{2\sqrt{1 + x}}=\frac{1}{8\sqrt{(1 + x)(1+\sqrt{1 + x})(1+\sqrt{1+\sqrt{1 + x}})}}$.

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