During our pretty extensive correcting session, we also decided that besides knowing that
To check on our progress with Part 2 of the exploratory, we did some work with logarithm functions and their inverses in Geogebra.
So, our function was
Note the two points on the graph... (1,5) and (3,6). The first point can be found by taking %3Dlog_31%2B5)
. Remember that %3Dlog_31)
is the same as saying "What power do you raise 3 to to get 1?" Any number raised to zero is 1 so the log (at any base) of one is always zero. 0+5 is 5, so we have the first point. Doing the same thing, (what power do you raise 3 to to get 3?), you find the point (3,6).
Now, to graph the inverse function, we'll use our set of steps...
1. Replace f(x) with y
2. Replace the Ys with Xs and vice versa.
3. Solve for y.
If you work through it, %3D3%5E%7Bx-5%7D%20)
When we graph it, we already know two points on the graph. When graphed, the inverse of a function looks like the graph of the original reflected across the line y=x, so when we reverse the x and y values of the points on f(x) we get points: (6,3) and (5,1) on g(x).
To make sure we knew how to find the derivative of a logarithm, we next found the derivate of f(x) at point (3,6).
so,
Note: On the calculator portion of an exam, you can hit "2nd-->trace" to get to the calc function, then hit "dy/dx" for a quick deriv. value.
So, now we know a point on the line, and the slope at that particular point, so we know the equation for the line. )
Surely enough when graphed, the line is tangent to the function.
For the line tangent to g(x) at (6,3), we have two options. One is to use the derivative rule for exponents, but the other way is far simpler. We know that f(x) and g(x) are inverses of each other, so we know that the two inverse points have reciprocal slopes. Therefore the equation of line tangent to g(x) at (6,3), is )
.
So, with our various derivative rules, we had only one kind of function we could not take the derivative of: Inverse Trig Functions.
As some background information, we discussed that the inverse functions only exist in certain quadrants of the unit circle. Arcsin and arctan exist in quadrants four and one, while the arccos function, exists in the first and second quadrants. We also discussed that trig functions equal ratios (like 
) while inverse trig functions equal angles (like 45 degrees). For a quick review of important inverse trigonometric info, here's some help from wolframalpha:
Using the example problem cos(arcsin(1/4)), we remembered how to compose trigonometric functions. We know that arcsin is an angle, so we can say that: )
and therefore 
. This is where our trigonometric properties from last year come in handy. Remember that 
? Well, it does, so we can use it to find cos(arcsin(1\4)). so, 
. Solve for cosine of the angle, and we get: 
.
Using this very tactic, we finally have everything we need to solve for the derivatives of inverse trigonometric functions...
In class, we started with y=arcsin(x), which we know is the same as sin(y)=x. Take the derivative with respect to x, and we get: 
or 
. Now, we can solve for cos(y) since we know that . We've already stated that sin(y)=x so, 
, and 
. Plug that back in for cos(y) in our derivative equation and we have the following rule:
Amazingly, it took implicit differentiation, the chain rule, and trigonometric rules to find the derivative of arcsin. For arccos, we went through a similar process:
so...
Two down...
And that's when we ran out of time... tune in next time for how to find the derivative of arctan... (or just look on http://www.wolframalpha.com/input/?i=derivative+of+arctan). Hint... it has to do with a certain other trigonometric rule...
The next scribe is Julia.
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