What happens if you can write a function as a composite in...

What happens if you can write a function as a composite in different ways? Do you get the same derivative each time? The Chain Rule says you should. Try it with the functions in Exercises 91 and 92. Find $d y / d x$ if $y=x^{3 / 2}$ by using the Chain Rule with $y$ as a composite of a. $y=u^{3} \quad$ and $\quad u=\sqrt{x}$ b. $y=\sqrt{u}$ and $u=x^{3}$

What happens if you can write a function as a composite in different ways? Do you get the same derivative each time? The Chain Rule says you should. Try it with the functions in Exercises 91 and 92.
Find $d y / d x$ if $y=x^{3 / 2}$ by using the Chain Rule with $y$ as a composite of
a. $y=u^{3} \quad$ and $\quad u=\sqrt{x}$
b. $y=\sqrt{u}$ and $u=x^{3}$

Key Concepts

Invariance Under Different Decompositions

When a function can be expressed as a composition of functions in multiple ways, each valid decomposition will yield the same derivative when the chain rule is applied correctly. This reflects the fact that the derivative is an intrinsic property of the function itself, independent of how it is decomposed mathematically. It reinforces the consistency and reliability of differentiation techniques in calculus.

Chain Rule

The chain rule is a fundamental principle in calculus used to compute the derivative of a composite function. It states that if a function is composed of two or more functions, the derivative of the composite function with respect to the original variable is the product of the derivative of the outer function evaluated at the inner function and the derivative of the inner function. This concept is crucial because it allows for systematic differentiation even when functions are nested within each other.

Composite Functions

Composite functions occur when one function is applied to the result of another function. In the context of differentiation, recognizing a composite function is important because it allows the use of the chain rule. Understanding composite functions helps in breaking down complex expressions into simpler parts, each of which can be differentiated on its own before being recombined to obtain the overall derivative.

Power Rule

The power rule is a basic rule for differentiating functions of the form x^n, where n is a real number. It states that the derivative of x^n is n*x^(n-1). This rule is often used in conjunction with the chain rule when the inner or outer function in a composition is a power function, thereby simplifying the differentiation process.

Transcript

00:01 All right, so we are finding the derivative of y equals x to the three halves power via the chain rule.

00:08 And we're actually going to do it via two different compositions, y equals u cubed and u equals square of x.

00:13 And then kind of the opposite, y equals square root of x, are squared of u and u equals x cubed.

00:21 Now this problem is yet again just checking out if the chain rule works.

00:25 So what they really want you to do is to show that both of these derivatives in a and b give you the same derivative.

00:31 As what we would get here via just directly deriving it, which in this case is three halves x to the one half.

00:43 Or if you want to write it as square root of x, you can, three half square of x.

00:48 Okay, let's check if the chain rule works, and if both of these compositions give us three halves x to the one half.

00:57 Now just to make my point abundantly clear, notice that if we just plug in for you right here, what we'll get is, y equals square root of x cubed which we can write as x to the one half cubed which ultimately is x to the three halves so we're just showing that the chain rule works and we can do something similar with part b and i encourage you to do so all right so they want us to find derivatives via the chain rule so let's go ahead and do just that okay we can definitely find a derivative of the derivative of y with respect to u 3 u squared which since we're going to want stuff in terms of x is just 3 squared of x squared or just 3x and we can also find the derivative of u with respect to x which is just going to be um and by the way we should probably write x as our square of x is x to the one half what we get is one half x to the negative one half or one over two square root of x so just to be abundantly clear what i did right here is i said hey you is just x to the one all right but now in order to find our total derivative d y over d x hopefully i have enough room here that's just d -y over d -u, which is 3x, times d -u over d -x, which is just 1 over 2 square root of x.

03:14 Let's get that in blue...

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