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

Lesson 31: Implicit differentiation (advanced examples)

# 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.
• 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.
• 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?
• 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.
• Say you had a constant in front of the ln x. Would you just multiply the derivative of ln x by the constant?
• 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
• i understand all that you said. so what would be the derivative of log( ln x)?
• 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] )
• what is the first derivative of y=e^-x ln^x
• 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! :)
• Would the same process be applied to a variable that is raised to the natural log, such as y= x^lnx ?
• 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)]``
• How does e^lnx simplify to x?
• ln(x) is defined as the inverse function of e^x. If you compose inverse functions, you get back the original input, x.
• 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?
• 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))
• Can I get a blackhole badge?