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

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, $u = x^{2}\sin(x)$, so $u^\prime=2x\sin(x)+x^{2}\cos(x)$ by the product - rule ($(uv)^\prime = u^\prime v+uv^\prime$ where $u = x^{2}$ and $v=\sin(x)$), and $v = 1 + x^{2}$, $v^\prime=2x$. 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: Apply product - rule and chain - rule for (b)

Let $u=\sin^{2}(3x)$ and $v = \sin(4x^{5})$. First, for $u=\sin^{2}(3x)$, let $t=\sin(3x)$, then $u = t^{2}$. By the chain - rule, $\frac{du}{dx}=\frac{du}{dt}\cdot\frac{dt}{dx}$. $\frac{du}{dt}=2t$ and $\frac{dt}{dx}=3\cos(3x)$, so $\frac{du}{dx}=2\sin(3x)\cdot3\cos(3x)=3\sin(6x)$ (using the double - angle formula $\sin(2\alpha)=2\sin(\alpha)\cos(\alpha)$). For $v=\sin(4x^{5})$, by the chain - rule, $v^\prime = 20x^{4}\cos(4x^{5})$. Then $f^\prime(x)=u^\prime v+uv^\prime=3\sin(6x)\sin(4x^{5})+20x^{4}\cos(4x^{5})\sin^{2}(3x)$.

Step3: Apply chain - rule for (c)

Let $y = f(x)=\sqrt{1+\sqrt{1+\sqrt{1 + x}}}$, let $u = 1+\sqrt{1+\sqrt{1 + x}}$, so $y=\sqrt{u}$. Then $\frac{dy}{du}=\frac{1}{2\sqrt{u}}$. Let $v = 1+\sqrt{1 + x}$, then $u = 1+\sqrt{v}$, $\frac{du}{dv}=\frac{1}{2\sqrt{v}}$, and $\frac{dv}{dx}=\frac{1}{2\sqrt{1 + x}}$. By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dv}\cdot\frac{dv}{dx}=\frac{1}{2\sqrt{1+\sqrt{1+\sqrt{1 + x}}}}\cdot\frac{1}{2\sqrt{1+\sqrt{1 + x}}}\cdot\frac{1}{2\sqrt{1 + x}}=\frac{1}{8\sqrt{1 + x}\sqrt{1+\sqrt{1 + x}}\sqrt{1+\sqrt{1+\sqrt{1 + x}}}}$.

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