展开 5((d^3)/(dt^3)(t^3))(1-sin(4t))

解答

展开

5 (\frac{d ^{3}}{d t ^{3}} (t^{3})) (1 - sin (4 t))

5 \frac{d ^{3}}{d t ^{3}} (t^{3}) - 5 sin (4 t) \frac{d ^{3}}{d t ^{3}} (t^{3})

求解步骤

5 (\frac{d ^{3}}{d t ^{3}} (t^{3})) (1 - sin (4 t))

使用分配律:

a (b - c) = ab - a c

a = 5 (\frac{d ^{3}}{d t ^{3}} (t^{3})), b = 1, c = sin (4 t)

= 5 (\frac{d ^{3}}{d t ^{3}} (t^{3})) \cdot 1 - 5 (\frac{d ^{3}}{d t ^{3}} (t^{3})) sin (4 t)

= 5 \cdot 1 \cdot (\frac{d ^{3}}{d t ^{3}} (t^{3})) - 5 sin (4 t) (\frac{d ^{3}}{d t ^{3}} (t^{3}))

化简

5 \cdot 1 \cdot (\frac{d ^{3}}{d t ^{3}} (t^{3})) - 5 sin (4 t) (\frac{d ^{3}}{d t ^{3}} (t^{3})) : 5 \frac{d ^{3}}{d t ^{3}} (t^{3}) - 5 sin (4 t) \frac{d ^{3}}{d t ^{3}} (t^{3})

= 5 \frac{d ^{3}}{d t ^{3}} (t^{3}) - 5 sin (4 t) \frac{d ^{3}}{d t ^{3}} (t^{3})

e x p an d \frac{3}{4} (2 x + 10 = 2 (3 - 2 x) + 7)

e x p an d \frac{n ( 5 n - 7 )}{( n + 1 ) ^{2}}

e x p an d \frac{K}{s ^{2} \cdot ( s t + 1 )}

e x p an d (e^{t c o s (x)} (t) - e^{t s i n (x)} (t))^{2}

e x p an d 2 ((3, - 4, 5, 0, 0) - 5 (4, 8, - 2, - 1, 3) + (- 4, 7, - 5))