problem 1. differentiate the following functions. (a) 4pts. f(x)=x²sin(x)/(1 + x²) (b) 4pts…

Explanation:

Step1: Recall 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, $u = x^{2}\sin(x)$ 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 $u^\prime = 2x$ and $v^\prime=\cos(x)$, so $u^\prime=2x\sin(x)+x^{2}\cos(x)$. And $v^\prime = 2x$. Then $f^\prime(x)=\frac{(2x\sin(x)+x^{2}\cos(x))(1 + x^{2})-x^{2}\sin(x)\cdot2x}{(1 + x^{2})^{2}}=\frac{2x\sin(x)+2x^{3}\sin(x)+x^{2}\cos(x)+x^{4}\cos(x)-2x^{3}\sin(x)}{(1 + x^{2})^{2}}=\frac{2x\sin(x)+x^{2}\cos(x)+x^{4}\cos(x)}{(1 + x^{2})^{2}}$.

Step2: Recall product - rule for (b)

The product - rule states that if $y = u\cdot v$, 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 $u^\prime = 2\sin(3x)\cdot3\cos(3x)=3\sin(6x)$. 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: Recall chain - rule for (c)

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

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