problem 1. differentiate the following functions. (a) 4pts. f(x)=x^2*sin(x)/(1 + x^2) (b) 4pts…

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 $u = x^{2}$, $u^\prime = 2x$ and $v=\sin(x)$, $v^\prime=\cos(x)$), 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 $y = u\cdot v$, then $y^\prime=u^\prime v+uv^\prime$. Let $u=\sin^{2}(3x)$ and $v = \sin(4x^{5})$. First, for $u=\sin^{2}(3x)$, using the chain - rule: let $t=\sin(3x)$, then $u = t^{2}$, $u^\prime = 2t\cdot t^\prime=2\sin(3x)\cdot3\cos(3x)=3\sin(6x)$ (using the double - angle formula $\sin(2\alpha)=2\sin\alpha\cos\alpha$). And $v^\prime=\cos(4x^{5})\cdot20x^{4}$. Then $f^\prime(x)=3\sin(6x)\sin(4x^{5})+20x^{4}\cos(4x^{5})\sin^{2}(3x)$.

Step3: Apply chain - rule for (c)

Let $y = f(x)=\sqrt{1+\sqrt{1+\sqrt{1 + x}}}$, let $u = 1+\sqrt{1+\sqrt{1 + x}}$, $y=\sqrt{u}=u^{\frac{1}{2}}$, $y^\prime=\frac{1}{2}u^{-\frac{1}{2}}\cdot u^\prime$. Let $v = 1+\sqrt{1 + x}$, $u = 1 + v^{\frac{1}{2}}$, $u^\prime=\frac{1}{2}v^{-\frac{1}{2}}\cdot v^\prime$. Let $w=1 + x$, $v = 1+w^{\frac{1}{2}}$, $v^\prime=\frac{1}{2}w^{-\frac{1}{2}}\cdot w^\prime$. Since $w^\prime = 1$, then $v^\prime=\frac{1}{2\sqrt{1 + x}}$, $u^\prime=\frac{1}{2\sqrt{1+\sqrt{1 + x}}}\cdot\frac{1}{2\sqrt{1 + x}}$, 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) $f^\prime(x)=3\sin(6x)\sin(4x^{5})+20x^{4}\cos(4x^{5})\sin^{2}(3x)$ (c) $f^\prime(x)=\frac{1}{8\sqrt{1 + x}\sqrt{1+\sqrt{1 + x}}\sqrt{1+\sqrt{1+\sqrt{1 + x}}}}$