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}(3 x) sin left(4 x^{5}
ight) )
(c) 4pts. ( f(x)=sqrt{1+sqrt{1+sqrt{1+x}}} )

Explanation:

Step1: Apply quotient rule

Let $u = x^2\sin x$, $v = 1+x^2$. Quotient rule: $\frac{u'v - uv'}{v^2}$.
$u' = 2x\sin x + x^2\cos x$, $v' = 2x$.
Expression: $\frac{(2x\sin x + x^2\cos x)(1+x^2) - x^2\sin x \cdot 2x}{(1+x^2)^2}$
Simplify numerator: $2x\sin x(1+x^2 - x^2) + x^2\cos x(1+x^2) = 2x\sin x + x^2(1+x^2)\cos x$

Step2: Apply product & chain rules

Let $u = \sin^2(3x)$, $v = \sin(4x^5)$. Product rule: $u'v + uv'$.
$u' = 2\sin(3x)\cdot 3\cos(3x) = 6\sin(3x)\cos(3x)$, $v' = \cos(4x^5)\cdot 20x^4 = 20x^4\cos(4x^5)$.
Expression: $6\sin(3x)\cos(3x)\sin(4x^5) + \sin^2(3x)\cdot 20x^4\cos(4x^5)$

Step3: Apply chain rule repeatedly

Rewrite $f(x) = [1 + \sqrt{1 + (1+x)^{1/2}}]^{1/2}$.
First derivative: $\frac{1}{2}[1 + \sqrt{1 + \sqrt{1+x}}]^{-1/2} \cdot \frac{d}{dx}[1 + \sqrt{1 + \sqrt{1+x}}]$
Second derivative inside: $\frac{1}{2}[1 + \sqrt{1+x}]^{-1/2} \cdot \frac{d}{dx}[1 + \sqrt{1+x}]$
Third derivative inside: $\frac{1}{2}(1+x)^{-1/2}$
Combine: $\frac{1}{2}[1 + \sqrt{1 + \sqrt{1+x}}]^{-1/2} \cdot \frac{1}{2}[1 + \sqrt{1+x}]^{-1/2} \cdot \frac{1}{2}(1+x)^{-1/2}$

Answer:

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