If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains ***.kastatic.org** and ***.kasandbox.org** are unblocked.

Main content

Current time:0:00Total duration:6:45

AP.CALC:

FUN‑3 (EU)

, FUN‑3.C (LO)

, FUN‑3.C.1 (EK)

so we have here f of X being equal to the natural log of the square root of x and what we want to do in this video is find the derivative of f and the key here is to recognize that f it can actually be viewed as a composition of two functions and we can diagram that out what's going on here well if you input an X into our function f what's the first thing that you do well you take the square root of it so if you if we start off with some X you input it the first thing that you do is you take the square root of it so you are going to take the square root of the input to produce the square root of x and then what do you do well you take the square root and then you take the natural log of that and so then you take the natural log of that so you could be you could view that as in putting it into another function that takes the natural log of whatever is inputted in I'm making these little squares to show what you do with the input and then what do you produce well you produce the natural log of the square root of x natural log of the square root of x which is equal to f of X so you could view f of X as this entire as this entire set or this entire I guess you could say this combination of functions right over there that is f of X which is essentially a composition of two functions you're inputting into one function then taking that output in putting it into another so you could have a function U here which takes the square root of whatever its input in is so U of X is equal to the square root of x and then you take another and then you take that output and input into another function that we could call V and what is V do well it takes a natural log of whatever the input is in this case in the case of F or in the case of where how I've just diagrammed it V is taking the natural log the input happens to be square root of x so it outputs the natural log of the square root of x if we wanted to write V with X as an input we would just say well that's just the natural log that is just the natural log of X and as you can see here f of X and I color-coded it ahead of time is equal to f of X is equal to the natural log of the square root of x so that is V of the square root of x or V of U of X so it is a composition which tells you that ok if I'm trying to find the derivative here the chain rule is going to be very very very very useful and the chain rule tells us that F prime of X is going to be equal to the derivative of you can view it as the outside function with respect to this inside function so it's going to be V prime of U of X V prime of U of x times the derivative of this inside function with respect to X so that's just u prime u prime of X so how do we evaluate these things well we know how to take the derivative of U of X and V of X u prime of X here is going to be equal to well remember square root of X is just the same thing as X to the one-half power so we can use the power rule bring the one-half out front so it becomes one-half X to the and then take off one of out of that exponent so that's one half minus one is negative one half power and what is V of X alright what is V prime of X well the derivative of the natural log of X is one over X we show that in other videos and so we now know what u prime of X is what is we know what V prime of X is but what is V prime of U of X well V prime of U of X wherever we see the X we replace it let me write that a little bit neater we replace that with a U of X so V prime of U of X is going to be equal to is going to be equal to one over U of X one over U of X which is equal to which is equal to one over U of X is just the square root of x 1 over the square root of x so this thing right over here we have figured out is one over the square root of x and this thing you prime of X we figured out is 1/2 times X to the negative 1/2 and X to the negative 1/2 I could rewrite that as 1/2 times 1 over X to the 1/2 which is the same thing as 1/2 times 1 over the square root of x or I could write that as 1 over 2 square roots of X so what is this thing going to be well this is going to be equal to in green V prime of U of X is 1 over the square root of x x times u prime of X is 1 over 2 times the square root of x now what is this going to be equal to well this is going to be equal to this is just algebra at this point 1 over we have our 2 and square root of x times square root of X is just X so it just simplifies to 1 over 2x so hopefully this made sense I intentionally diagrammed it out so you start to get that muscle in your brain going of recognizing the composite functions and then making a little bit more sense of some of these expressions of the chain rule that you might see in your calculus class or in your calculus textbook but as you get more practice you'll start to you'll be able to do it essentially without having to write out all of this you'll say ok look I have a composition this is the natural log of the square root of x this is V of U of X so what I want to do is I want to take the derivative of this outside function with respect to this inside function so the derivative of natural log of something with respect to that something is 1 over that something so it is 1 over that something the derivative of natural log of something with respect to that something is 1 over that something so that's what we just did here one way to think about it what was natural what would natural log of X be well that would be 1 over X but it's not natural log of X it's 1 over square root of x so it's going to be 1 over the square root of x so you take the derivative of the outside function with respect to the inside 1 and then you multiply that times just the derivative of side function with respect to X and we are done