0000:00:00:00:00Start12Number 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−2cosx
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−2cosx
