# if y=(4x+1)(1-x)^k where k is a constant find dy/dx?

• if y=(4x+1)(1-x)^k where k is a constant find dy/dx?

Answer #1 | 07/02 2014 05:34
here is a step by step solution: http://www.symbolab.com/solver/derivative-calculator/%5Cfrac%7Bd%7D%7Bdx%7D((4x%2B1)(1-x)%5E%7Bk%7D) hope it helps
Answer #2 | 07/02 2014 05:51
y=(4x+1)(1-x)^k y' = 4(1-x)^k - (4x+1)k(1-x)^(k-1) = (1-x)^(k-1)[4(1-x) - k(4x+1)] = (1-x)^(k-1)[4 - 4x - 4kx - k] = (1-x)^(k-1)[4 - k - 4(1 + k)x]
Answer #3 | 07/02 2014 03:32
To find the derivative we must use the product rule and chain rule... it is so dy/dx = 4(1-x)^k + (4x+1)*(-k)(1-x)^(k-1) OK!
Answer #4 | 07/02 2014 03:15
y=(4x+1)(1-x)^k diff by product rule w r t x dy/dx = (4x+1).(-k )(1 - x)^(k-1)+ 4(1 - x)^k...........................answer
Answer #5 | 07/02 2014 03:13
using chain rule d / dx((1-x)^k) = (1 - x)^(k - 1) d/dx(1 - x) = -(1 - x)^(k - 1) for whole expression use product rule (uv) ' = uv' + u'v d/dx[(4x+1)(1-x)^k] = (4x + 1)[ -(1 - x)^(k - 1)] + 4 (1-x)^k = 4 (1-x)^k - (4x + 1)(1 - x)^(k - 1) = (1 - x) ^(k-1) [ 4 (1-x) - (4x + 1)] = (1 - x) ^(k-1) [ 4 - 4x - 4x - 1] = (1 - x) ^(k-1) [ 3 - 8x]

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