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X
We're now going to use the inverse function theorem to figure out what the derivative
of the arc tangent of X ought to be. So the inverse function to arc tangent of X, would
be X equals tangent of Y. Now that's something we already know how to take the derivative
of. DX DY then, should be secant squared of Y. What does that mean? According to the theorem
over here, if we want DY DX, that should be 1 over DX DY. DY DX ought to equal 1 over
secant squared of Y. That's great and easy and everything but the problem is, we need
X's and we have Y's. This is where all these inverse functions get interesting. Especially
when they are trig functions. We need to put back in here, instead of a Y, an expression
in terms of X. So we know that X is equal to tangent of Y. We're going to use the fact
that tangent squared of Y is equal to secant squared of Y minus 1. That means that secant
squared of Y is equal to tangent squared of Y plus 1. We can replace the secant squared
of Y here in the bottom, with tangent squared plus 1. Now we're close to being able to go
to terms of X because we now that X is equal to tangent of Y. We get 1 over X squared plus
Y, replacing tangent of Y with X. That's the derivative of arc tangent. Now we can use
the same method to take the derivative of natural log of X. We've taken this derivative
in a previous video using straight forward implicit differentiation but now we can make
it a little bit quicker by using the inverse function theorem. If our function is natural
log of X, then its inverse is X equals E to the Y. That means that when we take the derivative,
we get DX DY is equal to the derivative of E to the Y, E to the Y. According to the theorem
over here, DY DX is 1 over E to the Y. We know that E to the Y here is X. The derivative
of natural log of X should be 1 over X, which is exactly what we found it to be in a previous
video. Now we're going to use this inverse function theorem to find the equation to the
tangent line of F inverse of Y, when X is 1. Even if we don't solve our actual equation
for F inverse. What do we mean? According to this part down here about a particular
point, we want the equation of the tangent line of F inverse of Y when X is 1. First
we better figure out what Y is. F of 1 is 5 times 1 squared, plus 1 equals 6. Our Y
value is going to be 6. 1 is our X value. Whenever you are trying to find a tangent
line, you need both a slope and a point. We've got our point. Now we need a slope. Slope
we're looking for is going to be F inverse, prime, when Y is 6. F inverse prime, when
Y is 6, is supposed to be the same as the derivative of X with respect to Y, evaluated
at Y equaling 6, which is supposed to be the same as 1 over the derivative of Y with respect
to X evaluated at X equaling 1, which is going to be 1 over F prime of 1. What's F prime
of 1? Well that's 1 over 10 X plus 1, evaluated when X is 1, which is 1 over 10 times 1, plus
1, which is one-eleventh. Now we have our slope. With our point and our slope, we can
find the equation of the tangent line. Now this is where it gets a little bit weird.
We're working with the inverse function of Y, in this particular case, we have X as a
function of Y, instead of Y as a function of X. Our tangent line, our point slope form
of the line, is going to actually have to be X minus the X value, equals M times Y minus
the Y value. We're going to get X minus the X value of 1, equals 1 over 11, Y minus the
Y value of 6.