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### Course: Class 12 math (India)>Unit 6

Lesson 7: Logarithmic differentiation

# Composite exponential function differentiation

Sal finds the derivatives of xˣ and x^(xˣ). They are surprisingly fun to do! Created by Sal Khan.

## Want to join the conversation?

• Why do you take the natural log of both sides at about "", instead of the regular log? Just wondering.
• I think there's some confusion here. log{to the base ten}(x) is actually just ln(x)/ln(10), where ln(x) is, of course, the natural log, i.e., to the base 'e', of x. I guess Sal has a video explaining that somewhere.

Anyway, ln(10) is just a constant. So it's very easy to see the proof:
d/dx{log(x)}
= d/dx{ln(x)/ln(10)}
= [d/dx{ln(x)}]/ln(10)
= 1/{x*ln(10)}

which is NOT the same thing as 1/{x*log(10)} , because ln(10) and log(10) are indeed different; log(10), in our context, being simply 1.

The key thing is the difference in the bases of the logarithms. 'e' happens to be an insanely cool number, which is why d/dx{ln(x)} = 1/x, while 10 isn't quite as cool, and thus d/dx{log(x)} = 1/{x*ln(10)}.
• at Sal call taking the 'x' from the power as a product of ln(x) like: x*ln(x). that is straight out of the logarithm properties, isn't it?!
• You are correct sir. Basic logarithm properties.
log (a^b) = b * log (a)
• In the second problem, how would you know the derivative of x^(x) without having done the first problem?
• I am a bit rusty, but I believe you could have taken the natural log of both sides again, or you could have just went through the product rule and then done the derivative of x^x as part of that.
• Does dy/dx = d/dx? If not, why?
• dy/dx means d/dx * (y). For most linear equations in the form y=f(x), this translates to d/dx * f(x). If y = f(x) + g(x) then dy/dx = d/dx * (y) = d/dx * f(x) + d/dx * g(x).
• Could you have started by redefining y = x^(x^x) as y = x^(x^2) ? Seems like it would be easier, but I haven't tried so...
• I think he meant the property about powers. When you are given `(x^a)^b`, then it is equal to `x^(a.b)`. So why not `x^(x^x) = x^(x.x) = x^(x^2)`? The answer is the use of parentheses.
• Can you differentiate this by using just the chain rule if you say that f(x)= x and g(x)= f(x)^x so it is a composite function?
• Great question! At first it seems like "of course you can!" But there is a big problem trying to use the chain rule on x^x to find the derivative of x^(x^x). Did you try it?
Think about how the chain rule usually works:
Given that `d/dx(sin(x)) = cos(x)`, I can say `d/dx(sin(u)) = cos(u)(du/dx)`, and in general:
Given that `d/dx(f(x)) = f'(x)`, I can say `d/dx(f(g(x))) = f'(g(x))g'(x)`
But when we start with `f(x) = x^x`, and `d/dx(f(x)) = f'(x) = x^x(ln(x) + 1)`, we run into a major problem that we don't actually know "which 'x' ", the base x or the exponent x, is where in the `f'(x) = x^x(ln(x) + 1)`. So you really need to do a bit more work if you want to use the chain rule. Basically, you need to start over, and find the derivative of f(x) = x^u, where u is some function of x, and you will find `d/dx(x^u) = x^u(ln(x)(du/dx) + u/x)`. So you find out, shockingly, that the 1 in the derivative was not really a 1! It was (exponent/base) which only becomes 1 when the exponent and base are both x! Armed with `d/dx(x^u) = x^u(ln(x)(du/dx) + u/x)`, you can now use the chain rule to find `d/dx(x^(x^x))`.
• so the power rule doesn't work if the exponent is a variable like 'x' right!
example: x^5 will work but not x^x at least not by the power rule.
• Correct. The power rule only works for constant exponents.
• What if the crazy problem gets crazier with the scaling of powers going on infinitely? I have come across a problem like that but with square roots.
• I am sorry but I hardly understand your explanation although the graph is very fascinating.
(1 vote)
• I don't get how to get the derivative of x^x. I came up with the answer of (ln x)x^x, but the answer is (lnx +1)x^x. How do you get that answer？ And how does x^x convert into e^(x lnx). Thanks to whomever can help me.