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## AP®︎ Calculus AB (2017 edition)

### Course: AP®︎ Calculus AB (2017 edition)>Unit 4

Lesson 8: Optional videos

# Derivative of ln(x) from derivative of 𝑒ˣ and implicit differentiation

How can you find the derivative of ln(x) by viewing it as the inverse of e^x? Created by Sal Khan.

## Want to join the conversation?

• Could I instead of replacing y=ln(x) on 1/(e^y), replace e^y=x on 1/(e^y)? Is there any problem when doing that? It seemed more simple than the way Sal did it.
(26 votes)
• There are usually about 42 ways of approaching a problem. The way you described is simpler, but sometimes Sal seems to enjoy the scenic route, so to speak.
(42 votes)
• Is there a video where he covers why the derivative of e^y is e^y * dy/dx? I thought it would just be e^y?
(11 votes)
• Well, worth remembering our y is a function of x. If you have y as a constant value (i know `y` wouldn't be the best letter to be used for a constant, but just in case), you can say that d(e^y)/dx = e^y.
(3 votes)
• Say you had a constant in front of the ln x. Would you just multiply the derivative of ln x by the constant?
(5 votes)
• Yes, that is correct. For example, d/dx of 5 ln x = 5/x

Note, however, that the d/dx of ln (ax) (where a is constant, not a variable) is a bit tricky.
d/dx ln (ax) = 1/x.
Here's proof:
y = ln (ax)
let u = ax
thus du/dx = a
y = ln u
dy/dx = d (ln u)/du ∙ du/dx
dy/dx = [1/u ] ∙ du/dx
back-substituting
dy/dx = [1/(ax)] ∙ a
dy/dx = 1/x
(18 votes)
• i understand all that you said. so what would be the derivative of log( ln x)?
(2 votes)
• Firstly log(ln x) has to be converted to the natural logarithm by the change of base formula as all formulas in calculus only work with logs with the base e and not 10.
Hence log( ln x ) = ln( ln x ) / ln (10) and then differentiating this gives [1/ln(10)] * [d(ln(ln x)) / dx].
This can be differentiated further by the Chain Rule, that gives [1/ln(10)]*{ [1/ln(x)*1/x ].
Hence the result is ( 1 / [ln(10)*ln(x)*x] )
(6 votes)
• what is the first derivative of y=e^-x ln^x
(3 votes)
• It's simple. You just need to know the rules. So first, take the first derivate of the entire thing. You'll get y' = (e^-x)' * (ln x) + (e^-x) * (ln x'). If you simplify this using derivative rules, you'll get y' = (e^-x * -1) * (ln x) + (e^-x) * (1/x).
Hope this helps! If you have any questions or need help, please ask! :)
(3 votes)
• Would the same process be applied to a variable that is raised to the natural log, such as y= x^lnx ?
(3 votes)
• That is a bit more complicated. Here is the derivative of that:
``y = x^ln xlet u = ln xthus, x = e^uthus, du = dx/xthus,y= x^ln x = (e^u)^u = e^(u²)Let w = u²Thus, dw = 2u due^(u²)= e^wTaking the derivative:dy = e^w dwdy = e^(u²) (2u)(du)dy = (x^ln x)(2 ln x)(dx/x)dy =  (2 ln x) x^[ln (x) - 1)]dxdy/dx =  (2 ln x) x^[ln (x) - 1)]``
(3 votes)
• How does e^lnx simplify to x?
(2 votes)
• ln(x) is defined as the inverse function of e^x. If you compose inverse functions, you get back the original input, x.
(4 votes)
• Can you use the chain rule to get the derivative of the natural log of the natural log of x? i.e ln(ln(x)). I saw this in a practice problem in composite exponential differentiation and tried to apply the chain rule, but the solution was wrong. So is the chain rule not applicable in this situation?
(2 votes)
• The chain rule applies in any situation where you have composed functions, and ln(ln(x)) is no exception. The chain rule gives you
d/dx ln(ln(x))
=1/(ln(x))·d/dx ln(x)
=1/(ln(x))·(1/x)
=1/(x·ln(x))
(3 votes)
• Can I get a blackhole badge?
(2 votes)
• you have to basically master all of Khan Academy to do that. XD
(1 vote)
• At about , why is d/dx of e^y = e^y * (dy/dx) ? I thought that by the power rule, d/dx should be y * e^(y-1) ... is it because d/dx is being taken of something other than x?
(1 vote)
• There are two reasons why what you said isn't true:
1) the derivative of e^x is e^x not xe^x-1
2) when your taking the derivative with respect to x of something that has a y you must apply the chain rule and take the derivative of the outer function (in this case e to the something.) with respect to that something. so you take d/dy of e^y first which gets you e^y, then you multiply by d/dx of the inner function which is y. we don't know what the derivative of y with respect to x is so we just write it as dy/dx, an unknown that you can solve for to figure out. so the answer is indeed e^y*dy/dx.
(2 votes)

## Video transcript

- [Voiceover] We know that the derivative with respect to x,of e to the x, is equal to e to the x, which I think is one of the neatest things in mathematics. It just builds my respect for the number e. But that's not what we're here to do, to just praise e. What we really want to think about is, what's the derivative of the inverse function? What's the derivative, with respect to x, of the natural log of x? We've done this several times, where we know the derivative of a function and we now want to find the derivative of the inverse function. What we can do, let's say y is equal to the natural log of x. This is another way of saying that y is the power that we raise e to, to get to x. So, this is equivalent to saying that e to the y power is equal to x. Now we can take the derivative of both sides of this equation with respect to x. So let's do that. The derivative of both sides with respect to x, do a little bit of implicit differentiation. Really just an application of the chain rule. So, on the left-hand side right over here, this is going to be the derivative of e to the y with respect to y, which is just going to be e to the y times the derivative of y with respect to x. On the right hand side here, the derivative of x with respect to x is one. Now, to solve for the derivative, we just divide both sides by e to the y. We get the derivative of y with respect to x is equal to one over e to the y power. Now, what is y equal to? Well, we know that y is equal to the natural log of x. So let's write that down. This is equal to one over e to the natural log of x. What is e to the natural log of x going to be? The natural log of x is the power I need to raise e to, to get to x. So if I actually raise e to that power, to that exponent, I'm going to get x. This is going to be equal to one over x. So this thing simplifies to x. We are done. We just figured out if y is equal to the natural log of x, the derivative of y with respect to x is one over x. Or you could say the derivative of the natural log of x with respect to x is equal to one over x. So this is equal to, this right over here, is equal to one over x. One over x, which is also a pretty neat result in mathematics. Not quite as exciting as this one, but still pretty neat.