problem 1.\ndifferentiate the following functions.\n(a) 4pts. $f(x)=\\frac{x^{2}\\sin(x)}{1 + x^{2}}$\n(b)…

Explanation:

Step1: Differentiate (f(x)=\frac{x^{2}\sin(x)}{1 + x^{2}}) using the quotient rule

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=x^{2}\cos(x)+2x\sin(x)) (using the 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).

[ \begin{align*} f^\prime(x)&=\frac{(x^{2}\cos(x)+2x\sin(x))(1 + x^{2})-x^{2}\sin(x)\cdot2x}{(1 + x^{2})^{2}}\ &=\frac{x^{2}\cos(x)+x^{4}\cos(x)+2x\sin(x)+2x^{3}\sin(x)-2x^{3}\sin(x)}{(1 + x^{2})^{2}}\ &=\frac{x^{2}\cos(x)+x^{4}\cos(x)+2x\sin(x)}{(1 + x^{2})^{2}} \end{align*} ]

Step2: Differentiate (f(x)=\sin^{2}(3x)\sin(4x^{5})) using the product rule and chain rule

The product rule: if (y = uv), (y^\prime=u^\prime v+uv^\prime). Let (u=\sin^{2}(3x)), (u^\prime = 2\sin(3x)\cos(3x)\cdot3=3\sin(6x)) (using the chain rule ((u^{n})^\prime=nu^{n - 1}u^\prime) with (u=\sin(3x)), (n = 2) and ((\sin(ax))^\prime=a\cos(ax))), and (v=\sin(4x^{5})), (v^\prime=\cos(4x^{5})\cdot20x^{4})

[ \begin{align*} f^\prime(x)&=3\sin(6x)\sin(4x^{5})+20x^{4}\cos(4x^{5})\sin^{2}(3x) \end{align*} ]

Step3: Differentiate (f(x)=\sqrt{1+\sqrt{1+\sqrt{1 + x}}}) using the chain rule

Let (y=\sqrt{u}), (u = 1+\sqrt{v}), (v=1+\sqrt{w}), (w = 1 + x).

(y^\prime=\frac{1}{2\sqrt{u}}u^\prime), (u^\prime=\frac{1}{2\sqrt{v}}v^\prime), (v^\prime=\frac{1}{2\sqrt{w}}w^\prime), (w^\prime=1)

[ \begin{align*} 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}} \end{align*} ]

Answer:

(a) (f^\prime(x)=\frac{x^{2}\cos(x)+x^{4}\cos(x)+2x\sin(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}})