How do you differentiate #f(x)=(4x+1)^2(1-x)^3# using the chain rule?

Question

Aurora Alvarado · Accepted Answer

#dy/dx=(4x+1)^2[-3(1-x)^2]+(1-x)^3(32x+8)#

Given - #y=(4x+1)^2(1-x)^3# #dy/dx=(4x+1)^2[3(1-x)^2(-1)]+(1-x)^3[2(4x+1)(4)]# #dy/dx=(4x+1)^2[-3(1-x)^2]+(1-x)^3(32x+8)#

Emily Ammerman · Answer

undefined To differentiate $ f(x) = (4x + 1)^2(1 - x)^3 $ using the chain rule, follow these steps:

1. Identify the inner functions and their derivatives:
   Let $ u = 4x + 1 $ and $ v = 1 - x $.
   Find $ \frac{du}{dx} $ and $ \frac{dv}{dx} $:
   $ \frac{du}{dx} = 4 $ and $ \frac{dv}{dx} = -1 $.

2. Apply the chain rule:
   $ \frac{df}{dx} = \frac{df}{du} \cdot \frac{du}{dx} + \frac{df}{dv} \cdot \frac{dv}{dx} $.

3. Find $ \frac{df}{du} $ and $ \frac{df}{dv} $:
   $ \frac{df}{du} = 2(4x + 1) \cdot 2(1 - x)^3 $ and $ \frac{df}{dv} = (4x + 1)^2 \cdot 3(1 - x)^2 $.

4. Substitute $ \frac{du}{dx} $, $ \frac{dv}{dx} $, $ \frac{df}{du} $, and $ \frac{df}{dv} $ into the chain rule formula:
   $ \frac{df}{dx} = (2(4x + 1) \cdot 2(1 - x)^3) \cdot 4 + ((4x + 1)^2 \cdot 3(1 - x)^2) \cdot (-1) $.

5. Simplify the expression:
   $ \frac{df}{dx} = 8(4x + 1)(1 - x)^3 - 3(4x + 1)^2(1 - x)^2 $.