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: Differentiate \(f(x)=\frac{x^{2}\sin(x)}{1 + x^{2}}\) using 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 \((uv)^\prime = u^\prime v+uv^\prime\) (where \(u = x^{2}\), \(u^\prime=2x\) and \(v=\sin(x)\), \(v^\prime=\cos(x)\)). So \(u^\prime=2x\sin(x)+x^{2}\cos(x)\). And \(v^\prime = 2x\).
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 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 \(u\), using the chain rule \((u^{n})^\prime=nu^{n - 1}u^\prime\) (here \(n = 2\), \(u=\sin(3x)\), \(u^\prime=3\cos(3x)\)), so \(u^\prime=2\sin(3x)\cdot3\cos(3x)=6\sin(3x)\cos(3x)\).
For \(v\), using the chain rule \((\sin(u))^\prime=\cos(u)u^\prime\) (here \(u = 4x^{5}\), \(u^\prime=20x^{4}\)), so \(v^\prime=20x^{4}\cos(4x^{5})\).
Then \(f^\prime(x)=6\sin(3x)\cos(3x)\sin(4x^{5})+20x^{4}\cos(4x^{5})\sin^{2}(3x)\)

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

Let \(y=\sqrt{u}\), \(u = 1+\sqrt{v}\), \(v=1+\sqrt{w}\), \(w = 1 + x\).
First, \(\frac{dy}{du}=\frac{1}{2\sqrt{u}}\), \(\frac{du}{dv}=\frac{1}{2\sqrt{v}}\), \(\frac{dv}{dw}=\frac{1}{2\sqrt{w}}\), \(\frac{dw}{dx}=1\).
By the chain rule \(\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dv}\cdot\frac{dv}{dw}\cdot\frac{dw}{dx}\)
\(\frac{dy}{dx}=\frac{1}{2\sqrt{1+\sqrt{1+\sqrt{1 + x}}}}\cdot\frac{1}{2\sqrt{1+\sqrt{1 + x}}}\cdot\frac{1}{2\sqrt{1 + x}}\)

Answer:

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