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Differential Calculus
Course: Differential Calculus > Unit 3
Lesson 11: Logarithmic differentiationWorked example: Composite exponential function differentiation
Sal differentiates the composite exponential function [ln(x)]ˣ and evaluates the derivative at x=e. Composite exponential functions are functions where the variable is in both the base and the exponent.
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I don't understand why the power x in the beginning can't be brought down in front of ln(x). Could somebody explain?(11 votes)
Notice how the power x is outside the brackets [lnx]^x and not [lnx^x]. The log property that allows you to bring the exponent down only applies in the second case, lnx^x. Thus why he needs to take ln on both sides to make the power inside the outer set of ln, which would allow him to apply the property.(26 votes)
Silly question, but were we not supposed to figure out d/dx, and ended up figuring out dy/dx - I thought the two were not equivalent?(3 votes)
d/dx is "change with respect to x." dy/dx is "change in y with respect to x." In general, dy/dx is used as a number describing how y changes with x, while d/dx is an operator which requires taking the derivative with respect to x. As such d/dx(y)=dy/dx.
This difference may seem a bit silly now, bit it will be very important when you deal with multiple interdependent functions. For example, if you have y(x), z(x), and x(t), then keeping track of dx/dt, dy/dx, dz/dx, dy/dt, and dz/dt will be very important.(12 votes)
- This can't be solved using...
y=[ln(x)]^x
power rule + chain rule -> x[ln(x)]^(x-1) * 1/x
But why? What rule does this break?(3 votes)
The power rule isn't applicable when there's an x in the exponent. When that's the case, you're actually looking at a variant of the exponential function (e^x), which behaves differently.
For completeness, I feel obligated to point out that the power rule only applies to polynomials.(8 votes)
At1:02, is there any particular reason he's using a natural log and not log base 10?(3 votes)
Differentiating natural log is a lot easier than differentiating log base 10 for L'Hopital's rule thing. Also math people just like natural log because 1. its about everyone's favourite number 'e', 2. its the original one, and 3. you don't have to wright the 'o' and the '10' and can write an 'n' instead of a 'g'(3 votes)
- could an alternate solution be this:
y={lnx]^x
raise both side to the 1/x power
y^1/x=lnx meaning that e^(y^1/x)=x
d/dx[e^(y^1/x)]=d/dx[x]
e^(y^1/x)*(y*(-1/x^2)+(1/x)*dy/dx)=1
-y/x^2+(1/x)*(dy/dx)=1/e^(y^1/x)
multiply both sides by x
-y/x+dy/dx=x/e^(y^1/x)
dy/dx=(-x^2)/x(e^(y^1/x))
and because y=ln(x)^2
dy/dx=(-x^2)/x(e^(lnx^1/x))
dy/dx=(-x^2)/x(e^lnx)
Thank You!(2 votes)
Unfortunately, your calculation produces non-real numbers for all negative x and -1 for all positive x, thus it doesn't appear to be an alternate solution.(1 vote)
applying the chain rule:
[ln(x)]^x
ln(ln(x))*[ln(x)]^x*(1/x)
ln(ln(e))*[ln(e)]^e*(1/e) = 0
is that not right?(1 vote)- It seems like what you're doing is applying the formula for [dy/dx] (a^x), and then applying the chain rule from there. The problem with this is that this only works when a is a constant. If a is another function you cannot use this formula.
In general, a composite function takes the form of f(g(x)); that is, g(x) replaces the x value. If g is instead replacing a constant, that isn't a composite function (at least, not a composite function with f and g!) but something else entirely. This means you cannot use the chain rule and need to find another approach.
Good thought though!(2 votes)
- I find it easier to write y = [ln(x)]^x = {e^[ln(ln(x))]}^x = e^[ln(ln(x)) * x], then differentiate using the chain rule.(1 vote)
I was good up until x * x^x is equal to x^x-1. I thought this would become 1^x. What am I missing here?(1 vote)
cant we solve this by applying chain rule and power rule?(1 vote)- I have a doubt. in the end, he imagined that if the question was : what is the derivative of y when x is equal to e? and the result was one. But, using the original equation , y= ln[x]^x , using x= e, we find y = 1. and dy/dx =0. Why this happened ? thank you(1 vote)
- Couldn't it be ln y=x ln^2x? At2:37?(1 vote)
1:02Video transcript
- [Voiceover] Let's say y is equal to the natural log of x to the xth power, and what we wanna do, we wanna find the derivative of y with respect to x. So I encourage you to pause this video and see if you can do it. So, when you first try to tackle this, this is a little bit daunting. We know how to take the
derivative of constants to some x power, but how
do we take a derivative of some time of a function, in this case, a natural log, to the xth power? And the answer here is to use some of our logarithmic properties,
and then we're going to do a little bit of implicit differentiation. So the first thing that we wanna do, actually, let me rewrite this
with a little bit of space, so, this was a natural
log of x to the xth power. So the first thing, I
wanna get rid of this x as an exponential, and I wanna be able to apply the product rule somehow. The way we're gonna do that is by taking the natural log of both sides. So take the natural log of both sides, and you might say, well,
why is that helpful? Well, if I'm taking the natural log of something to an exponent,
well, this is the same thing, actually, let me write this down as a property that you
may or may not remember from your logarithmic properties, so if I have, I could write log, or I'll just write natural log, of, if I have natural
log of a to the b power, this is the same thing as b
times the natural log of a. It's just a standard logarithmic property. And so, by taking the
natural log of both sides, this exponent can now become out front and scale the natural log function. So this exponent now, we
can bring that out front, and let's just rewrite everything. So we get the natural log of y is equal to, so then let
me put that in parentheses, so it's the natural log of y, is equal to x, that x is in blue, x times the natural log times
the natural log, of, sorry, x times the natural log
of the natural log of x. The natural log of the natural log of x. So, there you have it. By just taking the
natural log of both sides and using this logarithmic property, we were able to get that. Now you're saying, well, gee, how is this actually
going to be useful for us? Well now, we can implicitly take the derivative of both sides of this. And actually, let me scoot this over to the right a little
bit, just so that we can, I have space for my derivative operator. So there you go, scooted that over. And so now, let's take the derivative with respect to x of both sides. So let me, so I'm gonna
take the derivative with respect to x of the left-hand side and of the right-hand side, and of the right-hand side. Now, on the left-hand
side, this is going to be, essentially, an application
of the chain rule. When you learn implicit differentiation, it's really just application
of the chain rule. It's the derivative of
the outside function with respect to to the inside function, so the natural log of y with respect to y, the derivative of that is
just going to be one over y, one over y times the derivative of the inside function with respect to x. So dy dx, dy dx. That is going to be equal to, well this is going to get
interesting a little bit. Actually, let me do some stuff
on the side a little bit. Let me just, the first
thing we wanna do here is just apply the product rule. So it's the derivative
of the first expression, so it's just going to
be 1 times the second, I guess you'd say, function, so times the natural log
of the natural log of x, natural log of x, and then plus the first function, just x, times the derivative
of the second function. Times the derivative
of the second function. What's the derivative of the natural log of the natural log of x? Let's do that separately. So if I have, if I'm trying
to take the derivative, with respect to to x, of the natural log, the natural log of the natural log of x, of the natural log of x, well, here, again, I can
apply the chain rule. The derivative of that magenta function with respect to the inside function, that is going to be one
over the natural log of x, and then times the derivative
of the inside function with respect to x, so times one over x. So this is equal to one
over x natural log of x. So the derivative of this
second function right over here is one over x natural log of x, one over x natural log of x. Let's see, that x and that x cancels out, and so we are left with, we
are left with one over y, and I'll just write all of
this in this blue color, so one over y times the derivative of y with respect to x, is equal to, see this is just the natural
log of the natural log of x, the natural log of the natural log of x, plus one over the natural log of x, one over the natural log of x. And now, to solve for the derivative, we can multiply both sides
by y, so let's do that. So we're gonna multiply that side by y, and we're gonna multiply
this side times y, and what are we going to get? Well, on the left-hand side,
that's why we multiplied by y, we just have the derivative
y with respect to x, derivative of y with respect to x, is equal to, well, y is our original, is all this, is our
original thing that we had. Y was equal to the natural log, let me rewrite it over here. Y was equal to the natural
log of x to the xth power. So this is, we're essentially
multiplying both sides times the natural log
of x to the xth power. So this gonna get a little
bit, a little bit messy here. So we could just write it
the way I wrote it just now, without it being distributed, actually, let me just leave it like that. So it's going to be, and so we deserve a drum roll right now, 'cause
this is quite involved, the natural log of the natural log of x plus one over the natural log of x, all of that times the natural
log of x to the xth power. So that was quite involved. And if someone said well,
what is the derivative of y when x is equal to e? So if someone says what is this equal to when x is equal to e, well,
we could evaluate this when x is equal to e. This would be the, and I
just made that up just now, so if like the original
question wasn't just what is dy dx, if they said what is dy dx when x is equal to e. If that was the original question, then we could evaluate it. So where we just replace
all these with e's, so it'd be an e there,
an e there, an e there, and an e there, and I
just picked the value e 'cause it's easy to evaluate. So that natural log of e is 1, natural log of 1 e to
the zeroth power is 1, so all of that just becomes 0. The natural log of e is 1, so this whole expression right over here becomes zero plus one over one, so this becomes one, and then the natural log of e is,
the natural log of e is one, and you're gonna have
one to the eth power, well you could raise one to any power, and you're just get one. So it's one times one is equal to one. So I just thought it would be fun to try to evaluate that at a value that would be somewhat clean.