Question

1. $f(x)=2\pi^{4}$

$f(x)=8\pi^{3}$ or $f(x)=0$

1. $f(x)=\sqrt{\sin^{3}x}$

$f(x)=\frac{1}{2\sqrt{\sin^{3}x}}\frac{3\sin^{2}x\cos x}{}$ or $f(x)=\frac{1}{2\sqrt{3\sin^{2}x\cos x}}$

1. $f(x)=\cos(x^{3})$

$f(x)=-\sin(3x^{2})6x$ or $f(x)=-3x^{2}\sin(x^{3})$

1. $f(x)=x^{2}\cos x$

$f(x)=2x(-\sin x)$ or $f(x)=2x\cos x - x^{2}\sin x$

Explanation:

Step1: Differentiate \(f(x) = 2\pi^{4}\)

Since \(\pi\) is a constant, the derivative of a constant - multiple function \(y = cf(x)\) (where \(c\) is a constant) and the derivative of a constant \(y = k\) (\(k\) is a constant) are used. The derivative of a constant \(k\) is \(0\), and for \(y=2\pi^{4}\), \(f'(x)=0\).

Step2: Differentiate \(f(x)=\sqrt{\sin^{3}x}=(\sin x)^{\frac{3}{2}}\)

Use the chain - rule \(\frac{d}{dx}(u^{n})=nu^{n - 1}\cdot u'\), where \(u = \sin x\) and \(n=\frac{3}{2}\). First, \(\frac{d}{dx}(\sin x)^{\frac{3}{2}}=\frac{3}{2}(\sin x)^{\frac{3}{2}-1}\cdot\frac{d}{dx}(\sin x)\). Then, since \(\frac{d}{dx}(\sin x)=\cos x\), we have \(f'(x)=\frac{3}{2}\sqrt{\sin x}\cos x=\frac{3\sin x\cos x}{2\sqrt{\sin x}}=\frac{1}{2\sqrt{\sin^{3}x}}\cdot3\sin^{2}x\cos x\).

Step3: Differentiate \(f(x)=\cos(x^{3})\)

Use the chain - rule \(\frac{d}{dx}(f(g(x)))=f'(g(x))\cdot g'(x)\). Let \(g(x)=x^{3}\) and \(f(u)=\cos u\). Then \(f'(u)=-\sin u\) and \(g'(x) = 3x^{2}\). So \(f'(x)=-\sin(x^{3})\cdot3x^{2}=- 3x^{2}\sin(x^{3})\).

Step4: Differentiate \(f(x)=x^{2}\cos x\)

Use the product - rule \((uv)' = u'v+uv'\), where \(u = x^{2}\) and \(v=\cos x\). \(u' = 2x\) and \(v'=-\sin x\). Then \(f'(x)=2x\cos x+x^{2}(-\sin x)=2x\cos x - x^{2}\sin x\).

Answer:

1. \(f'(x)=0\)
2. \(f'(x)=\frac{3\sin^{2}x\cos x}{2\sqrt{\sin^{3}x}}\)
3. \(f'(x)=-3x^{2}\sin(x^{3})\)
4. \(f'(x)=2x\cos x - x^{2}\sin x\)