differentiate the following functions. (a) 4pts. f(x) = (x^2 sin(x))/(1 + x^2) (b) 4pts. f(x) = sin^2(3x)…

Answer:

(a) $f^\prime(x)=\frac{(2x\sin(x)+x^{2}\cos(x))(1 + x^{2})-2x\cdot x^{2}\sin(x)}{(1 + x^{2})^{2}}$ (b) $f^\prime(x)=2\sin(3x)\cos(3x)\cdot3\cdot\sin(4x^{5})+\sin^{2}(3x)\cdot\cos(4x^{5})\cdot20x^{4}$ (c) $f^\prime(x)=\frac{1}{2\sqrt{1+\sqrt{1+\sqrt{1 + x}}}}\cdot\frac{1}{2\sqrt{1+\sqrt{1 + x}}}\cdot\frac{1}{2\sqrt{1 + x}}$

Explanation:

Step1: Apply quotient - rule for (a)

Quotient rule: $(\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$, where $u = x^{2}\sin(x)$, $v=1 + x^{2}$. First find $u^\prime$ using product - rule: $u^\prime=(x^{2})^\prime\sin(x)+x^{2}(\sin(x))^\prime=2x\sin(x)+x^{2}\cos(x)$, $v^\prime = 2x$. Then $f^\prime(x)=\frac{(2x\sin(x)+x^{2}\cos(x))(1 + x^{2})-2x\cdot x^{2}\sin(x)}{(1 + x^{2})^{2}}$.

Step2: Apply product - rule for (b)

Product rule: $(uv)^\prime=u^\prime v+uv^\prime$, where $u=\sin^{2}(3x)$ and $v = \sin(4x^{5})$. For $u=\sin^{2}(3x)$, let $t=\sin(3x)$, then $u = t^{2}$, $u^\prime=2t\cdot t^\prime=2\sin(3x)\cos(3x)\cdot3$. $v^\prime=\cos(4x^{5})\cdot20x^{4}$. So $f^\prime(x)=2\sin(3x)\cos(3x)\cdot3\cdot\sin(4x^{5})+\sin^{2}(3x)\cdot\cos(4x^{5})\cdot20x^{4}$.

Step3: Apply chain - rule for (c)

Chain rule: If $y = f(g(x))$, then $y^\prime=f^\prime(g(x))\cdot g^\prime(x)$. Let $y=\sqrt{1+\sqrt{1+\sqrt{1 + x}}}$, let $u = 1+\sqrt{1+\sqrt{1 + x}}$, then $y=\sqrt{u}$, $y^\prime=\frac{1}{2\sqrt{u}}\cdot u^\prime$. Continuing to apply chain - rule for $u$ and its inner functions, we get $f^\prime(x)=\frac{1}{2\sqrt{1+\sqrt{1+\sqrt{1 + x}}}}\cdot\frac{1}{2\sqrt{1+\sqrt{1 + x}}}\cdot\frac{1}{2\sqrt{1 + x}}$.