00:01
Suppose you want to find an equation of the tangent line to the curve.
00:04
Why? which is equal to two? x plus one over x plus two.
00:08
At the .1, 1 to do this, we first find the slope of the tangent line.
00:14
That is the derivative of the function evaluated at the given .1, 1.
00:23
Now by quotient rule we have white prime, that's equal to x plus two times the derivative of the numerator two, x plus one-.
00:34
We have to express one times the derivative of the denominator which is expressed to this all over the square of the denominator.
00:45
And then from here we have x plus two times derivative of to express one.
00:51
That's just too minus two, x plus one times the derivative of x plus two which is one.
00:58
And then this all over the square of x plus two.
01:02
And simplifying this, we have two, x plus four minus two, x -1.
01:11
This all over x plus two squared or this is just three over the square of experts to and so the slope of the tangent line at the .11 is given by that's dy over dx evaluated at 1 1...