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Lecture

# 4.3 exponential.pdf

4 Pages
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School
University of Toronto St. George
Department
Mathematics
Course
MAT237Y1
Professor
Dan Dolderman
Semester
Fall

Description
Calc 1 Wed 10/8/08 4.3 - Derivatives of Exponential and Inverse Trig functions First let’s talk about inverse functions. Remember, in order for the inverse of a function to exist, the function must be one-to-one. Ie, it must pass the horizontal line test. So, say we have a one-to-one di▯erentiable function f. What other conditions are necessary in order foto be di▯erentiable? Think about how the inverse is a re ection over the line y = x. In order tor f be di▯erentiable, we don’t want it to have any vertical tangents. What would correspond to a vertical tangent when re ected over the line y = x? How about a horizontal line? Hmm... Okay, ▯1 so basically, in order for to exist and be di▯erentiable, f needs to be one-to-one and have no horizontal tangents. So now we must ask ourselves, \what is the derivative of the inverse of a function?" Let y = f▯1(x). Then f(y) = x. Now let’s do some implicit di▯erentiation. 0 dy f (y) = 1 dx And so dy = 1 dx f (y) But what’s y? Why, that’s just f1(x)! Therefore, dy 1 = dx f (f▯1(x)) Or, if you want to think of it this way, dy = 1 : dx dx=dy Example: let f(x) = x . Find the derivative of the inverse. x+1 y x = y + 1 x(y + 1) = y xy + x = y x = y ▯ xy x = y(1 ▯ x) x y = 1 ▯ x So f▯1(x) = 1▯x. And, well, calculating the derivative directly, ▯ ▯ d x = (1 ▯ x) ▯ x(▯1) dx 1 ▯ x (1 ▯ x) = 1 (x ▯ 1)2 1 Using the formula, we get (x + 1) ▯ x 1 f (x) = = (x + 1)2 (x + 1)2 So, 1 1 (x + (1 ▯ x)) 1 = = = f (f▯1 (x)) 1 2 (1 ▯ x)2 (x ▯ 1)2 ((1▯x)+1) Of course, it pretty much didn’t matter whether or not we took the derivative of the inverse directly or we used the formula, but we did need to solve for the inverse both times. Sometimes it’s a bit harder to solve the derivative of one over the other, so this formula may be useful at times. Let’s now go on to ▯nding the derivative of f(x) = e . x y = e ln(y) = xln(e) ln(y) = x 1 dy = 1 y dx dy = y dx dy = e x dx Oh, snap! The derivative of e is itself! That’s so cool! What’s even cooler is that we could have saved ourselves a little time back there by just using that derivative of the inverse thing! ▯1 x If f(x) = ln(x) then f (x) = e . So, d 1 e = = ex dx 1x (e ) Similarly, for any base b, d x x dx[b ] = b ln(b) Example: ▯nd d[ex +1] dx d x +1 x +1 [e ] = (2x)e dx Be careful, just because something is raised to some power
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