Question

Если f(x)=2sinx^2 * cosx, то чему равно f '(π/2)?

Answer 1

$f'(x) = (2\sin x^2)' \cdot \cos x + 2\sin x^2 \cdot (\cos x)' = 2\cos x^2 \cdot (x^2)' \cos x + 2 \sin x^2 \cdot (-\sin x) = 2\cos x^2 \cdot 2x \cos x - 2 \sin x^2 \cdot \sin x = 4x\cdot \cos x \cdot \cos x^2 - 2\sin x \cdot \sin x^2\\f'(\frac{\pi}{2}) = 4\cdot \frac{\pi}{2} \cdot \cos \frac{\pi}{2} \cos (\frac{\pi}{2})^2 - 2 \sin \frac{\pi}{2} \cdot \sin (\frac{\pi}{2})^2 = 2\pi\cdot 0\cdot \cos \frac{\pi^2}{4} - 2\cdot 1 \sin (\frac{\pi^2}{4}) = 0 - 2\sin (\frac{\pi^2}{4}) = - 2\sin (\frac{\pi^2}{4})$