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 ( (\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}} )

Let ( u = x^{2}\sin(x) ), ( v=1 + x^{2} ). First, find ( u^\prime ) using the product rule ( (uv)^\prime=u^\prime v+uv^\prime ). Let ( u_1=x^{2} ), ( v_1=\sin(x) ). Then ( u_1^\prime = 2x ), ( v_1^\prime=\cos(x) ). So ( u^\prime=2x\sin(x)+x^{2}\cos(x) ). And ( v^\prime = 2x ). [ \begin{align*} 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}} \end{align*} ]

Step2: Differentiate ( f(x)=\sin^{2}(3x)\sin(4x^{5}) ) using the product rule ( (uv)^\prime = u^\prime v+uv^\prime )

Let ( u=\sin^{2}(3x)=( \sin(3x))^{2} ), ( v = \sin(4x^{5}) ). For ( u ), using the chain rule ( (u_1^{n})^\prime=n u_1^{n - 1}u_1^\prime ), where ( u_1=\sin(3x) ), ( n = 2 ). Then ( u^\prime=2\sin(3x)\cos(3x)\cdot3=3\sin(6x) ) (since ( 2\sin\alpha\cos\alpha=\sin(2\alpha) )). For ( v ), using the chain rule, let ( t = 4x^{5} ), then ( 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 ( y = f(g(x))), ( y^\prime=f^\prime(g(x))\cdot g^\prime(x) )

Let ( y=\sqrt{u}=u^{\frac{1}{2}} ), ( u = 1+\sqrt{v}=1 + v^{\frac{1}{2}} ), ( v=1+\sqrt{w}=1+w^{\frac{1}{2}} ), ( w = 1 + x ). ( y^\prime=\frac{1}{2}u^{-\frac{1}{2}}\cdot u^\prime ), ( u^\prime=\frac{1}{2}v^{-\frac{1}{2}}\cdot v^\prime ), ( v^\prime=\frac{1}{2}w^{-\frac{1}{2}}\cdot w^\prime ), ( w^\prime = 1 ). [ \begin{align*} 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(1+\sqrt{1+\sqrt{1 + x}})^{\frac{1}{2}}(1+\sqrt{1 + x})^{\frac{1}{2}}(1 + x)^{\frac{1}{2}}} \end{align*} ]

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(1+\sqrt{1+\sqrt{1 + x}})^{\frac{1}{2}}(1+\sqrt{1 + x})^{\frac{1}{2}}(1 + x)^{\frac{1}{2}}} )