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 $v^\prime = 2x$. 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: 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 $v^\prime=\cos(4x^{5})\cdot20x^{4}$. Then, by the product - rule $f^\prime(x)=u^\prime v+uv^\prime=6\sin(3x)\cos(3x)\sin(4x^{5})+20x^{4}\sin^{2}(3x)\cos(4x^{5})$.

Step3: Recall chain - rule multiple times

Let $y = f(x)=\sqrt{1+\sqrt{1+\sqrt{1 + x}}}$, and let $u = 1+\sqrt{1+\sqrt{1 + x}}$, so $y=\sqrt{u}=u^{\frac{1}{2}}$, then $y^\prime=\frac{1}{2}u^{-\frac{1}{2}}\cdot u^\prime$. Let $v = 1+\sqrt{1 + x}$, then $u = 1 + v^{\frac{1}{2}}$, and $u^\prime=\frac{1}{2}v^{-\frac{1}{2}}\cdot v^\prime$. Let $w=1 + x$, then $v = 1+w^{\frac{1}{2}}$, and $v^\prime=\frac{1}{2}w^{-\frac{1}{2}}\cdot w^\prime$. Since $w^\prime = 1$, we have $v^\prime=\frac{1}{2\sqrt{1 + x}}$, $u^\prime=\frac{1}{2\sqrt{1+\sqrt{1 + x}}}\cdot\frac{1}{2\sqrt{1 + x}}$, and $y^\prime=\frac{1}{2\sqrt{1+\sqrt{1+\sqrt{1 + x}}}}\cdot\frac{1}{2\sqrt{1+\sqrt{1 + x}}}\cdot\frac{1}{2\sqrt{1 + x}}=\frac{1}{8\sqrt{1 + x}\sqrt{1+\sqrt{1 + x}}\sqrt{1+\sqrt{1+\sqrt{1 + x}}}}$.

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