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:

(a) Step1: Use the quotient rule

The quotient rule is \((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}\), where \(u = x^{2}\sin(x)\) and \(v=1 + x^{2}\).
First, find \(u^\prime\) using the product rule \((uv)^\prime = u^\prime v+uv^\prime\) (here \(u = x^{2}\), \(v=\sin(x)\)). So \(u^\prime=2x\sin(x)+x^{2}\cos(x)\), and \(v^\prime = 2x\).

(a) Step2: Substitute into the quotient rule

Substitute \(u\), \(u^\prime\), \(v\), \(v^\prime\) into \(\frac{u^\prime v - uv^\prime}{v^{2}}\), we get \(f^{\prime}(x)=\frac{(2x\sin(x)+x^{2}\cos(x))(1 + x^{2})-2x\cdot x^{2}\sin(x)}{(1 + x^{2})^{2}}\)

(b) Step1: Use the product rule and chain rule

The product rule \((uv)^\prime=u^\prime v + uv^\prime\), where \(u=\sin^{2}(3x)\) and \(v = \sin(4x^{5})\).
For \(u=\sin^{2}(3x)\), use the chain rule \(y = u^{2}\), \(u=\sin(t)\), \(t = 3x\). Then \(u^\prime=2\sin(3x)\cos(3x)\cdot3\).
For \(v=\sin(4x^{5})\), use the chain rule \(y=\sin(u)\), \(u = 4x^{5}\), then \(v^\prime=4\times5x^{4}\cos(4x^{5})\)

(b) Step2: Substitute into the product rule

Substitute \(u^\prime\) and \(v^\prime\) into \((uv)^\prime=u^\prime v+uv^\prime\), we have \(f^{\prime}(x)=6\sin(3x)\cos(3x)\sin(4x^{5})+20x^{4}\sin^{2}(3x)\cos(4x^{5})\)

(c) Step1: Use the chain rule multiple - times

Let \(y = \sqrt{u}\), \(u=1+\sqrt{v}\), \(v = 1+\sqrt{w}\), \(w=1 + x\).
First, \(y^\prime=\frac{1}{2\sqrt{u}}u^\prime\), \(u^\prime=\frac{1}{2\sqrt{v}}v^\prime\), \(v^\prime=\frac{1}{2\sqrt{w}}w^\prime\), and \(w^\prime = 1\)

(c) Step2: Substitute back

Substitute \(u\), \(v\), \(w\) and their derivatives: \(f^{\prime}(x)=\frac{1}{2\sqrt{1+\sqrt{1+\sqrt{1 + x}}}}\cdot\frac{1}{2\sqrt{1+\sqrt{1 + x}}}\cdot\frac{1}{2\sqrt{1 + x}}\)
Simplify to get \(f^{\prime}(x)=\frac{1}{8\sqrt{1 + \sqrt{1+\sqrt{1 + x}}}\cdot\sqrt{1+\sqrt{1 + x}}\cdot\sqrt{1 + x}}\)

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)

$$f^{\prime}(x)=6\sin(3x)\cos(3x)\sin(4x^{5})+20x^{4}\sin^{2}(3x)\cos(4x^{5})$$

(c)

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