differentiate the functions with respect to x.
y=(x+1)^2(3x+2)^2
Ans:2(x+1)(3x+2)(6x+5)
A.Maths
2007-05-12 7:41 am
回答 (3)
2007-05-12 7:59 am
✔ 最佳答案
y=(x+1)^2(3x+2)^2y'=[(x+1)^2] d/dx [(3x+2)^2] + [(3x+2)^2] d/dx [(x+1)^2]...............(by Product Rule)
=[(x+1)^2]*[2(3x+2) d/dx(3x+2)] + [(3x+2)^2]*[2(x+1) d/dx(x+1)].........(by Chain Rule)
=[(x+1)^2]*[2(3x+2)*3] + [(3x+2)^2]*[2(x+1)] ............................ (Simplify)
=[(x+1)^2]*[6(3x+2)] + [(3x+2)^2]*2(x+1)
=(x+1)(3x+2)[6(x+1) + 2(3x+2)]
=2(x+1)(3x+2)[3(x+1) + (3x+2)]
=2(x+1)(3x+2)(6x+5)
2007-05-12 7:53 am
y' = 2(x+1)(1) (3x+2)^2 + (x+1)^2 (2)(3x+2)(3)
= 2(x+1) (3x+2)^2 + 6(x+1)^2(3x+2)
= 2(x+1)(3x+2) [ (3x+2) + 3(x+1) ]
= 2 (x+1) (3x+2) (6x+5)
= 2(x+1) (3x+2)^2 + 6(x+1)^2(3x+2)
= 2(x+1)(3x+2) [ (3x+2) + 3(x+1) ]
= 2 (x+1) (3x+2) (6x+5)
2007-05-12 7:51 am
y = (x + 1)^2 (3x + 2)^2
dy/dx = 2(x + 1)(1)(3x + 2)^2 + 2(3x + 2)(3)(x + 1)^2 ...... (Product Rule)
= 2(x + 1)(3x + 2)[(3x + 2) + 3(x + 1)] ......(Factorization)
= 2(x + 1)(3x + 2)(6x + 5)
dy/dx = 2(x + 1)(1)(3x + 2)^2 + 2(3x + 2)(3)(x + 1)^2 ...... (Product Rule)
= 2(x + 1)(3x + 2)[(3x + 2) + 3(x + 1)] ......(Factorization)
= 2(x + 1)(3x + 2)(6x + 5)
收錄日期: 2021-05-03 06:58:53
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