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: Use the quotient rule for (a)

The quotient rule is \((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^2}\). Let \(u = x^{2}\sin(x)\), \(u^\prime=2x\sin(x)+x^{2}\cos(x)\) (using product rule \((uv)^\prime = u^\prime v+uv^\prime\) where \(u = x^{2}\), \(u^\prime=2x\) and \(v=\sin(x)\), \(v^\prime=\cos(x)\)), and \(v = 1 + x^{2}\), \(v^\prime=2x\). 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: Use the product and chain rules for (b)

The product rule \((uv)^\prime=u^\prime v + uv^\prime\). Let \(u=\sin^{2}(3x)\), using the chain rule \(u^\prime = 2\sin(3x)\cos(3x)\cdot3=3\sin(6x)\) (since \(\sin(2\alpha)=2\sin\alpha\cos\alpha\)), and \(v = \sin(4x^{5})\), \(v^\prime=\cos(4x^{5})\cdot20x^{4}\). Then \(f^\prime(x)=3\sin(6x)\sin(4x^{5})+20x^{4}\cos(4x^{5})\sin^{2}(3x)\)

Step3: Use the chain rule multiple - times for (c)

Let \(y = f(x)=\sqrt{1+\sqrt{1+\sqrt{1 + x}}}\). Let \(u = 1+\sqrt{1+\sqrt{1 + x}}\), \(y=\sqrt{u}\), \(y^\prime=\frac{1}{2\sqrt{u}}\cdot u^\prime\). Let \(v = 1+\sqrt{1 + x}\), \(u = 1+\sqrt{v}\), \(u^\prime=\frac{1}{2\sqrt{v}}\cdot v^\prime\). And \(v^\prime=\frac{1}{2\sqrt{1 + x}}\). Then \(f^\prime(x)=\frac{1}{2\sqrt{1+\sqrt{1+\sqrt{1 + x}}}}\cdot\frac{1}{2\sqrt{1+\sqrt{1 + x}}}\cdot\frac{1}{2\sqrt{1 + x}}=\frac{1}{8\sqrt{1 + x}\sqrt{1+\sqrt{1 + x}}\sqrt{1+\sqrt{1+\sqrt{1 + x}}}}\)

Answer:

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