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: Apply the quotient rule

The quotient rule states that if \(y=\frac{u}{v}\), then \(y^{\prime}=\frac{u^{\prime}v - uv^{\prime}}{v^{2}}\). Here, \(u = x^{2}\sin(x)\) and \(v=1 + x^{2}\).
First, find \(u^{\prime}\) using the product rule (\((ab)^{\prime}=a^{\prime}b+ab^{\prime}\), where \(a = x^{2}\), \(a^{\prime}=2x\), \(b=\sin(x)\), \(b^{\prime}=\cos(x)\)). So \(u^{\prime}=2x\sin(x)+x^{2}\cos(x)\), and \(v^{\prime}=2x\).

Step2: Substitute into the quotient rule formula

$$
LATEXBLOCK0
$$

(b)

Step1: Apply the product rule

The product rule \(y = ab\), \(y^{\prime}=a^{\prime}b+ab^{\prime}\), where \(a=\sin^{2}(3x)\) and \(b = \sin(4x^{5})\).
For \(a=\sin^{2}(3x)\), use the chain rule (\((u^{n})^{\prime}=nu^{n - 1}u^{\prime}\), let \(u=\sin(3x)\), \(u^{\prime}=3\cos(3x)\)). So \(a^{\prime}=2\sin(3x)\cdot3\cos(3x)=6\sin(3x)\cos(3x)\).
For \(b=\sin(4x^{5})\), use the chain rule (\((\sin(u))^{\prime}=\cos(u)u^{\prime}\), let \(u = 4x^{5}\), \(u^{\prime}=20x^{4}\)). So \(b^{\prime}=20x^{4}\cos(4x^{5})\).

Step2: Substitute into the product rule formula

$$
LATEXBLOCK1
$$

(c)

Step1: Let \(y=\sqrt{1 + \sqrt{1+\sqrt{1 + x}}}\), and use the chain rule multiple - times

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

Step2: Substitute back

$$
LATEXBLOCK2
$$

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)

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