Question

problem 1. differentiate the following functions. (a) 4pts. $f(x)=\frac{x^{2}\sin(x)}{1 + x^{2}}$ (b) 4pts. $f(x)=\sin^{2}(3x)\sin(4x^{5})$ (c) 4pts. $f(x)=\sqrt{1+\sqrt{1+\sqrt{1 + x}}}$

Explanation:

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

The quotient rule is \((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}\), where \(u = x^{2}\sin(x)\), \(v=1 + x^{2}\)
First, find \(u^\prime\) using product rule \((uv)^\prime=u^\prime v+uv^\prime\) (here \(u = x^{2}\), \(v=\sin(x)\)), so \(u^\prime=2x\sin(x)+x^{2}\cos(x)\)
\(v^\prime = 2x\)
Then \(f^\prime(x)=\frac{(2x\sin(x)+x^{2}\cos(x))(1 + x^{2})-x^{2}\sin(x)\cdot2x}{(1 + x^{2})^{2}}=\frac{2x\sin(x)+2x^{3}\sin(x)+x^{2}\cos(x)+x^{4}\cos(x)-2x^{3}\sin(x)}{(1 + x^{2})^{2}}=\frac{2x\sin(x)+x^{2}\cos(x)+x^{4}\cos(x)}{(1 + x^{2})^{2}}\)

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

The product rule \((uv)^\prime=u^\prime v+uv^\prime\), let \(u=\sin^{2}(3x)\), \(v = \sin(4x^{5})\)
For \(u\), using chain rule: let \(t=\sin(3x)\), \(u=t^{2}\), \(u^\prime=2t\cdot3\cos(3x)=6\sin(3x)\cos(3x)\)
For \(v\), using chain rule: let \(s = 4x^{5}\), \(v=\sin(s)\), \(v^\prime=20x^{4}\cos(4x^{5})\)
\(f^\prime(x)=6\sin(3x)\cos(3x)\sin(4x^{5})+\sin^{2}(3x)\cdot20x^{4}\cos(4x^{5})\)

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

Let \(y=\sqrt{1+\sqrt{1+\sqrt{1 + x}}}\), let \(u = 1+\sqrt{1+\sqrt{1 + x}}\), \(y=\sqrt{u}\), \(y^\prime=\frac{1}{2\sqrt{u}}\cdot u^\prime\)
Let \(v=\sqrt{1+\sqrt{1 + x}}\), \(u = 1 + v\), \(u^\prime=\frac{1}{2\sqrt{v}}\cdot v^\prime\)
Let \(w=\sqrt{1 + x}\), \(v=\sqrt{1+w}\), \(v^\prime=\frac{1}{2\sqrt{1 + w}}\cdot w^\prime\)
\(w^\prime=\frac{1}{2\sqrt{1 + x}}\)
Back - substitute: \(v^\prime=\frac{1}{2\sqrt{1+\sqrt{1 + x}}}\cdot\frac{1}{2\sqrt{1 + x}}\), \(u^\prime=\frac{1}{2\sqrt{\sqrt{1+\sqrt{1 + x}}}}\cdot\frac{1}{2\sqrt{1+\sqrt{1 + x}}}\cdot\frac{1}{2\sqrt{1 + x}}\)
\(f^\prime(x)=\frac{1}{2\sqrt{1+\sqrt{1+\sqrt{1 + x}}}}\cdot\frac{1}{2\sqrt{1+\sqrt{1 + x}}}\cdot\frac{1}{2\sqrt{1 + x}}\)

Answer:

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