if ( f(x)=cos (sin (x^{2})) ), find ( f^{prime}(x) ).\nfind ( f^{prime}(2) ).

Explanation:

Step1: Apply the chain rule

The chain rule states that if (y = f(g(x))), then (y'=f'(g(x))\cdot g'(x)). Let (u = \sin(x^{2})), so (f(x)=\cos(u)). Then (f'(x)=-\sin(u)\cdot u').

Step2: Differentiate (u = \sin(x^{2}))

Let (v=x^{2}), so (u = \sin(v)). By the chain rule, (u'=\cos(v)\cdot v'). Since (v = x^{2}), (v'=2x). So (u'=\cos(x^{2})\cdot2x).

Step3: Substitute (u) and (u') back

(f'(x)=-\sin(\sin(x^{2}))\cdot\cos(x^{2})\cdot2x=- 2x\cos(x^{2})\sin(\sin(x^{2})))

Step4: Find (f'(2))

Substitute (x = 2) into (f'(x)): (f'(2)=-2\times2\times\cos(2^{2})\times\sin(\sin(2^{2}))=-4\cos(4)\sin(\sin(4)))

Using a calculator (in radian mode): (\cos(4)\approx - 0.6536), (\sin(4)\approx - 0.7568), (\sin(\sin(4))\approx\sin(-0.7568)\approx - 0.6856)

(f'(2)=-4\times(-0.6536)\times(-0.6856)\approx-4\times0.4483\approx - 1.793)

Answer:

(f'(x)=-2x\cos(x^{2})\sin(\sin(x^{2}))) and (f'(2)\approx - 1.793)