Calculus AB Unit 3

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21 Terms

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Chain Rule


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Implicit Differentiation means/requires what?

When you differentiate in terms of y (so anytime y is involved), you must multiply dy/dx to the value containing y. Then to fully differentiate, get all dy/dx alone on one side.

<p>When you differentiate in terms of y (so anytime y is involved), you must multiply dy/dx to the value containing y. Then to fully differentiate, get all dy/dx alone on one side.</p>
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How to find the Derivative at a point (with implicit differentiation)

Differentiate given equation, then solve for dy/dx (your slope).

Separately, calculate the point (sometimes already given, sometimes said as, “when x=#”) that you will be using for point-slope.

Plug in the point values to your dy/dx so you can get a non-variable slope.

Then use that value and the point to make a point-slope form equation for the tangent line. (Sometimes there’s more than one point, so there may be more than one line.)

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Finding (all) Horizontal Tangent Lines

Differentiate equation, solving for dy/dx.

Horizontal tangent lines exist when the slope (dy/dx) = 0

So set your dy/dx value equal to 0 and solve for x.

Then plug that x value into your original f(x) to get your y=### line(s).

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Finding (all) Vertical Tangent Lines

Differentiate equation, solving for dy/dx.

Vertical tangent lines exist when the slope (dy/dx) = undefined.

So set your dy/dx value equal to undefined and find what y value will make it true.

Then plug that y value into your original f(x) to get your x=### line(s).

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Finding the Inverse of a Function

To find a function’s inverse, swap the input (x) and output (y) values.

<p>To find a function’s inverse, swap the input (x) and output (y) values.</p>
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Derivative of an Inverse Function

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When simplifying inverse sec and inverse csc derivatives, what do you have to check?

Test x=-1 in your expression to see which sign the numerator and denominator turn out to be. If it started negative, then you must include or not include the absolute value in the denominator to ensure the simplification retains that same sign pattern.

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Domain (and range) of inverse sin (on unit circle)

Right Quadrants I and IV

<p>Right Quadrants I and IV</p>
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Domain (and range) of inverse cos (on unit circle)

Top Quadrants I and II

<p>Top Quadrants I and II</p>
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Domain (and range) of inverse tan (on unit circle)

Right Quadrants I and IV (excluding pi and 3pi/2)

<p>Right Quadrants I and IV (excluding pi and 3pi/2)</p>
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1st Derivative Notation

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2nd Derivative Notation

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nth Derivative Notation

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When you’re taking a second derivative and have a dy/dx in your answer, then you can…

substitute it for the first derivative.