Post #...1/10/2010

Instead of balantly typing the steps for the first and second derivatives on here, I have decided to do a couple of problems from the AP exams and explain them in terms I'm pretty sure you will understand as well (although this remains to be seen).

The following problem falls into the LINEARIZATION category:

**When the local linearization of f(x) = sqrt(9 + sin2x) near 0 is used, an estimate of 0.06 is...

Linearization is just the equation of a tangent line with the x value they give you plugged in.

f'(x) = cos2x / sqrt(9 + sin2x)

Since youknow that your x value is 0 (near 0), plug it in to find your slope.

This then gives you 1/sqrt(9), whcih simplifies to 1/3.
This is your slope of the tangent line. Now the only thing you are missing is your y value. You find this by plugging your x value into your original which gives you the point (0,3)

So now you have a point and a slope..plug in to point slope form.

y-3 =1/3(x-0)

Now plug in your 0.06 (since you are approximating to that) into your equation of tangent line. This then yields 3.02


Now for an easy problem:

If y = sin^3(1-2x), then dy/dx is...

As you can tell, the problem is just telling you to take the derivative. Now this part may be a little tricky because it is indeed a chain rule that involves multiple steps, thus possibly having a negative effect on your final answer.

Ok. so first deal witht the exponent:

dy/dx = 3sin^2(1-2x)

Now the derivative of sin:

dy/dx = 3sin^2(1-2x)cos(1-2x)

Now the derivative of the inside:

dy/dx = 3sin^2(1-2x)cos(1-2x)(-2)

Which then simplifies further to:

dy/dx = -6sin^2(1-2x)cos(1-2x)


Ok. Thats enough examples for now..will do more in the future!!

~mal