00:01
Okay, so we're given the equation of a line, and we're asked to find the tangent to the curve at x equals 1.
00:10
So, keep in mind you need two things to write the equation of a line.
00:15
You need a point and a slope.
00:19
So they've given us the x coordinate of the point.
00:23
To find the y coordinate of the point, we just need to go into the original function.
00:28
So let's plug in 1 for x.
00:32
And i get 2 times the tangent of pi times 1 is just pi.
00:41
So the tangent of pi over 4 is 1, and 2 times 1 gives me a y value of 2.
00:49
So i know that my tangent line goes through the point 1 -2.
00:55
Now, to find the slope, i do the derivative.
00:59
So i'll have to use chain rule.
01:03
So the scalar says i'm going to take the derivative of the out.
01:08
Outside function.
01:09
So the derivative of tangent is secant squared.
01:14
I'm going to freeze the inside function, so pi x over four stays the same, times the derivative of the inside function, which would be pi over four...