Help with a derivative word problem
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Help with a derivative word problem

[From: ] [author: ] [Date: 12-06-25] [Hit: ]
For the second part, recall that percentage change is (dx/x) or (dy/y), in other words, the incremental change in a quantity divide by the quantity itself.When x=2 we have y=5,dy = (6x-2)dx,......
The function y is definedby y= 3x^2 -2x -3. Write down it's derivative with respect to x. Calculate the approximate percentage change in y when x=2 and x increases by p%.

I got the derivative: 6x-2, how do I go about doing the next steps?
Pls explain

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You've got the correct expression for the derivative.

For the second part, recall that percentage change is (dx/x) or (dy/y), in other words, the incremental change in a quantity divide by the quantity itself. When x=2 we have y=5, so

dy = (6x-2)dx, which at x=2 means dy = 10dx

Now rewrite this as

(dy/5) = 4 (dx/2)

i.e., (dy/y) = 4 (dx/x)

Thus when x changes by p percent, y will change by 4p percent (for small values of p).

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The reason that y changes by (4p) percent is that (dy/y) is four times (dx/x) at the point x=2 and y=5.

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