Determine the equation of the tangent to the curve y=(x+1)^2(x-1) at (2, 9)? y = 15x -21 y = 15x - 39 y =-15x -21 y = 15x +39
Added by Carla B.
Step 1
This will give us the slope of the tangent line at any point on the curve. Using the product rule, we have: \[ y' = \frac{d}{dx}[(x+1)^2(x-1)] \] \[ y' = (x+1)^2 \frac{d}{dx}(x-1) + (x-1) \frac{d}{dx}(x+1)^2 \] \[ y' = (x+1)^2(1) + (x-1)(2(x+1)) \] \[ y' = Show more…
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