12Number 12ExplanationGiven f(x)=sin2xf(x) = \sin^2 xf(x)=sin2x. If f′(x)f'(x)f′(x) represents the first derivative of f(x)f(x)f(x), then limh→∞h[f′(x+1h)−f′(x)]=....\lim_{h \to \infty} h \left[f'\left(x + \frac{1}{h}\right) - f'(x)\right] = ....h→∞limh[f′(x+h1)−f′(x)]=....sin2x\sin 2xsin2x−cos2x-\cos 2x−cos2x2cos2x2 \cos 2x2cos2x2sinx2 \sin x2sinx−2cosx-2 \cos x−2cosxExplanationGiven f(x)=sin2xf(x) = \sin^2 xf(x)=sin2x Asked limh→∞h[f′(x+1h)−f′(x)]=...?\lim_{h \to \infty} h \left[f'\left(x + \frac{1}{h}\right) - f'(x)\right] = ... ?h→∞limh[f′(x+h1)−f′(x)]=...? Let 1h=a\frac{1}{h} = ah1=a; h→∞h \to \inftyh→∞ so the value a→0a \to 0a→0. limh→∞h[f′(x+1h)−f′(x)]\lim_{h \to \infty} h \left[f'\left(x + \frac{1}{h}\right) - f'(x)\right]h→∞limh[f′(x+h1)−f′(x)]limh→∞f′(x+a)−f′(x)a\lim_{h \to \infty} \frac{f'(x + a) - f'(x)}{a}h→∞limaf′(x+a)−f′(x)f′′(x)f''(x)f′′(x) Since f(x)=sin2xf(x) = \sin^2 xf(x)=sin2x, then f(x)=sin2xf(x) = \sin^2 xf(x)=sin2xf′(x)=2sinxcosxf'(x) = 2 \sin x \cos xf′(x)=2sinxcosxf′′(x)=2cos2xf''(x) = 2 \cos 2xf′′(x)=2cos2x Therefore, the value of limh→∞h[f′(x+1h)−f′(x)]=2cos2x\lim_{h \to \infty} h \left[f'\left(x + \frac{1}{h}\right) - f'(x)\right] = 2 \cos 2xh→∞limh[f′(x+h1)−f′(x)]=2cos2x
