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I'm in an introductory Calculus class and would really like to understand $u$-substitution. So far, I have been able to understand all the concepts but hit a brick wall here. I know that in $u$-substitution, you have $u\cdot\frac{du}{dx}$ and you set whatever $u$ is, well, equal to $u$. However, why is it that you then derive $u$ to get $\frac{du}{dx} = \text{something}$ and must solve for $dx$ and plug in? That's the part that got me lost. I understand what $u$ should be set to, but after that I'm unsure about what actually happens. Could someone guide me (simply) through the process and why $u$-substitution works like this?

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  • $\begingroup$ Think of it as the "reverse chain-rule", just like how integration by parts is like "reverse product-rule." You've seen a proof I hope of why $[f(g(x))]' = f'(g(x))g'(x)$. Using the fundamental theorem of calculus, you can get from the one to the other. See proof-wiki's proof. As for the second half of your question, that is exactly what the chain rule tells you to do, $\frac{d}{dx}[\ln(u)]=\frac{d}{du}[\ln(u)]\cdot\frac{d}{dx}[u] = \frac{1}{u}\frac{du}{dx}$ $\endgroup$
    – JMoravitz
    Mar 22, 2015 at 19:56
  • $\begingroup$ Hi, thank you so much for your reply. Could you direct me to a good, easy to understand proof of the chain rule? I have not found one online. Thank you for your reply and explanation.(I did not realize it was chain rule, my mistake) $\endgroup$
    – MW130
    Mar 22, 2015 at 20:10
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    $\begingroup$ I don't know why people call it "u-substitution." What if I substitute it a different variable? Just call it substitution. $\endgroup$ Mar 22, 2015 at 20:36
  • $\begingroup$ There is a nice section about replacing dx on Insight Things. It seems to be a good idea to think about u-substitution as replacing $x$ in terms of $u$. Some ideas are more comprehensible then. $\endgroup$ Apr 5, 2016 at 7:43

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Theorem 1: Derivative of composite functions (I.e. Chain-Rule):

When $y=f(u)$ is a differentiable function w.r.t $u$ and $u=g(x)$ is a differentiable function w.r.t $x$, then $y=f(g(x))$ is a differentiable function w.r.t $x$ and:

$\frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx}~~~~$

(note, these are not fractions and do not necessarily follow the same properties of real numbers. It is by convenient notation and properties of derivatives that this is true. I.e., you should not have expected that you can cancel the du's on top and bottom.)

Equivalently, by relabling and using different notation you have:

$\frac{d}{dx}[f(g(x))] = f'(g(x))\cdot g'(x)$, or equivalently yet, that for $u=g(x)$ you have

$\frac{d}{dx}[f(u)] = \frac{d}{du}[f(u)]\frac{du}{dx}$

Proof:

Let $h(x)=f(g(x))$. We wish to prove then that $h'(c) = f'(g(c))g'(c)$. Assume for a moment that $g(x)\neq g(c)$ in the neighborhood of $c$ (to avoid division by zero errors).

By the definition of derivatives:

$$\begin{align} h'(c) &= \lim\limits_{x\to c} \frac{h(x)-h(c)}{x-c}\\ &= \lim\limits_{x\to c} \frac{f(g(x))-f(g(c))}{x-c}\\ &= \lim\limits_{x\to c} \frac{f(g(x))-f(g(c))}{x-c}\cdot\frac{g(x)-g(c)}{g(x)-g(c)}\\ &= \lim\limits_{x\to c} \frac{f(g(x))-f(g(c))}{g(x)-g(c)}\cdot\frac{g(x)-g(c)}{x-c}\\ &= \left[\lim\limits_{x\to c} \frac{f(g(x))-f(g(c))}{g(x)-g(c)}\right] \cdot \left[\lim\limits_{x\to c} \frac{g(x)-g(c)}{x-c}\right]\\ &= f'(g(c))g'(c) \end{align} $$

In a more complete proof, we may choose to be a bet more precise for the case where $g(x)=g(c)$ within the neighborhood of $c$. Either it will be constant in the neighborhood of $c$, or we can always pick a small enough neighborhood such that you avoid the issue entirely (else it will contradict the differentiability of $g$)

Theorem 2: Integration by substitution:

Let $g$ be a function whose range is an interval $I$, and let $f$ be a function continuous on $I$. If $g$ is differentiable on its domain and $F$ is an antiderivative of $f$ on $I$, then:

$\int f(g(x))g'(x)dx = F(g(x))+C$

By relabling, setting $u=g(x)$, then $du=g'(x)dx$ and:

$\int f(u)du = F(u)+C$

Proof:

Let $y=F(u)$ and $u=g(x)$. Then by theorem 1 (chain-rule) above we get:

$\frac{d}{dx}[F(g(x))] = F'(g(x))g'(x)$.

By definition of antiderivatives then, $\int F'(g(x))g'(x)dx = F(g(x))+C = F(u)+C$

Since $F$ is an antiderivative of $f$, the result follows.

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    $\begingroup$ I cannot tell you how grateful I am for this post. This helped me so much - enlightened me on both theorems. Thank you SO much. Most helpful post I've ever seen on here. $\endgroup$
    – MW130
    Mar 22, 2015 at 21:00
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    $\begingroup$ As for examples and putting the theorems into practice, Might I recommend PatrickJMT's Youtube Channel $\endgroup$
    – JMoravitz
    Mar 22, 2015 at 21:01
  • $\begingroup$ Absolutely! Thanks again for your extremely well thought out and clear response! $\endgroup$
    – MW130
    Mar 22, 2015 at 21:02

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