differentiate the following functions.\n(a) 4pts. $f(x)=\frac{x^{2}sin(x)}{1 + x^{2}}$\n(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 $u^\prime = 2x\sin(x)+x^{2}\cos(x)$ 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 (b)

The product - rule: if $y = u\cdot v$, then $y^\prime=u^\prime v+uv^\prime$. Let $u=\sin^{2}(3x)$ and $v = \sin(4x^{5})$. For $u=\sin^{2}(3x)$, using the chain - rule, let $t=\sin(3x)$, then $u = t^{2}$, so $u^\prime=2\sin(3x)\cdot\cos(3x)\cdot3 = 3\sin(6x)$ (using the double - angle formula $\sin(2\alpha)=2\sin\alpha\cos\alpha$). For $v=\sin(4x^{5})$, $v^\prime=\cos(4x^{5})\cdot20x^{4}$. 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 $y = f(x)=\sqrt{1+\sqrt{1+\sqrt{1 + x}}}$, let $u = 1+\sqrt{1+\sqrt{1 + x}}$, so $y=\sqrt{u}=u^{\frac{1}{2}}$, $y^\prime=\frac{1}{2}u^{-\frac{1}{2}}\cdot u^\prime$. Let $v = 1+\sqrt{1 + x}$, then $u = 1 + \sqrt{v}$, $u^\prime=\frac{1}{2}v^{-\frac{1}{2}}\cdot v^\prime$. And $v^\prime=\frac{1}{2}(1 + x)^{-\frac{1}{2}}$. So $f^\prime(x)=\frac{1}{2}(1+\sqrt{1+\sqrt{1 + x}})^{-\frac{1}{2}}\cdot\frac{1}{2}(1+\sqrt{1 + x})^{-\frac{1}{2}}\cdot\frac{1}{2}(1 + x)^{-\frac{1}{2}}=\frac{1}{8\sqrt{1+\sqrt{1+\sqrt{1 + x}}}\sqrt{1+\sqrt{1 + x}}\sqrt{1 + x}}$.

Answer:

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