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### Course: Calculus, all content (2017 edition)>Unit 2

Lesson 36: Logarithmic differentiation

# Worked 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.

## Want to join the conversation?

• I don't understand why the power x in the beginning can't be brought down in front of ln(x). Could somebody explain?
• 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.
• 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?
• 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.
• 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?
• 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.
• At , is there any particular reason he's using a natural log and not log base 10?
• 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'
• 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!
• 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.
• 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?
• 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!
• 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.
• 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?
• I just came across a sum in my assignments where we need to find dy/dx from the equation y^y=sinx. Can I solve it the same way as depicted in this video?
• yln(y) = ln(sinx)

or

(e^ln(y))^y = sin(x)
• Why can't you just apply the chain rule right at the beginning with lnx as the inner function and [lnx]^x as the outer function?
• Unfortunately, you also have a ln (y) on the other side to be taken care of....and therefore, implicit differentiation is the way to go....however without that first move you could differentiate it explicitly.....but that would require plenty of hairy maths!