Class Notes
(808,660)

Canada
(493,333)

University of Toronto St. George
(42,810)

Mathematics
(2,710)

MAT237Y1
(52)

Dan Dolderman
(31)

Lecture

# 4.3 exponential.pdf

Unlock Document

University of Toronto St. George

Mathematics

MAT237Y1

Dan Dolderman

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

More
Less
Related notes for MAT237Y1