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## AP®︎ Calculus AB (2017 edition)

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

Lesson 7: Derivatives of logarithmic functions

# Worked example: Derivative of ln(√x) using the chain rule

In this worked example, we dissect the composite function f(x)=ln(√x) into its parts, ln(x) and √x. By applying the chain rule, we successfully differentiate this function, providing a clear step-by-step process for finding the derivative of similar composite functions.

## Want to join the conversation?

• This is a silly thing, but couldn't Sal have solved this without using the chain rule at all? We know from the power law of the natural log is that if the term inside it has an exponent (in this case, 1/2) we can bring it out. So that would leave us with 1/2*ln(x) which is just a million times simpler.
• Yes, you are correct. Sal was just trying to illustrate the chain rule.
• Is that notation correct? `sin' x` (sine prime of x)
• You mean the derivative of `sin(x)`?
It would be more correct to write it as
`d/dx sin(x)`
Or
`f'(x), f(x) = sin(x)`
I hope this helps!
• v'(u(x))= 1/u(x)... How did we get to this?
Because.... v'(x)= 1/(x)
So v'(u(x)) = 1/x rootx
• Let u(x)=p. Then we're taking the derivative with respect to p of v(p) is v'(p)=1/p.
Resubstitute p and we get v'=1/(√x).
By chain rule, to get the derivative of v with respect to x, we then multiply by u'(x). So our result is
1/(√x)•[1/(2√x)]=1/(2x).
• why is v'(u(x)) = 1/u(x) ??
• That is just in this case where the derivative of ln(x)* is *1/x , and v is representing *ln(x)*. So, *v'(x)* is just *1/x.*
• What if it is f(x)= ln(2x+5)^3, how wpuld the chain rule play into that, with the ^3 on the end?
• Where is the ^3 on?
If it's ln((2x+5)^3), then it equals to 3ln(2x+5), then f'(x)=3(ln(2x+5))'=3/(2x+5)*(2x+5)'=6/(2x+5)
If it's (ln(2x+5))^3, then the derivative equals to f'(x)= 3*(ln(2x+5))^2*(ln(2x+5))'=3*(ln(2x+5))^2*(1/(2x+5))*(2x+5)'=[6(ln(2x+5))^2]/(2x+5).
• Why use the chain rule here? Why not simplify to (ln x)/2 and differentiate that? = 1/2 d/dx(ln(x)) =(1/x)/2 = 1/2x. Much quicker.
• Good point. But this isn't always the case. in other cases using chain rule is faster or even it is the only way.
• At , why √x*√x equals x and not |x|?
• By definition, a square root of a number is any number that when multiplied by itself gives the original number.
• Could I write v'(u(x)) as (1/x)*sqrt x, and leave it like that?
• Lets look at the whole equation. It says that,
f(x)=ln(√x)
Using the chain rule, f'(x) is 1/(√x) (1/2√x) (1)
• Newton's notation seems to make it clear what any given derivative has been taken with respect to, since it sits on the denominator of the 'fraction'. Using Leibniz' notation, is it correct to say that whatever is in the function bracket defines what the derivative has been taken with respect to?
• I believe you have mixed up the notations. According to KA,
Newton's notation is ẏ or ḟ, most commonly used in physics.
This notation does not very clearly show what the derivative is with respect to.
Lagrange's notation is y’ or f’(x), pronounced "f prime".
The "x" in the brackets is what the derivative is wrt.
Leibniz's notation is the most common d/dx, df/dx, or dy/dx.
The "denominator" is the variable the derivative is wrt.
Hope that I helped, and correct me if I'm wrong.