**Implicit differentiation with the chain rule and in — Transcript & Summary | SozAI** Source: https://sozai.app/transcript/implicit-differentiation-chain-rule/ Brian McLogan explains implicit differentiation using the chain rule to find dy/dx for ln(x² + y²). ## Key Takeaways - Implicit differentiation requires applying the chain rule to functions involving y as a function of x. - Derivatives of composite functions like ln(x² + y²) involve differentiating inside terms carefully. - dy/dx must be isolated by grouping and factoring terms after differentiation. - Multiplying through by denominators can help clear fractions and simplify solving for dy/dx. - Understanding each step conceptually helps avoid common pitfalls in implicit differentiation. ## What the video covers - The video focuses on implicit differentiation of the function ln(x² + y²). - Brian McLogan applies the chain rule to differentiate both sides with respect to x. - He explains the derivative of the inside function, including derivatives of x² and y². - The derivative dy/dx is treated as an unknown to be solved for. - Steps include grouping terms with dy/dx on one side and factoring it out. - The video covers multiplying both sides by the denominator to simplify the expression. - Brian addresses common challenges in isolating dy/dx during implicit differentiation. - The final expression for dy/dx is derived by algebraic manipulation and factoring. - The explanation emphasizes understanding the chain rule in the context of implicit functions. - The video is a step-by-step walkthrough aimed at learners needing clarity on implicit differentiation. ## Chapters 1. 00:00 Introduction to implicit differentiation 2. 00:19 Setting up the derivative of ln(x² + y²) 3. 00:48 Applying the chain rule to inside functions 4. 01:44 Differentiating x² and y² terms 5. 02:18 Grouping terms and isolating dy/dx 6. 03:41 Factoring dy/dx and solving algebraically 7. 04:51 Final steps to express dy/dx explicitly 8. 05:11 Summary and review of the chain rule application Answers ## Questions about this video What is implicit differentiation? Implicit differentiation is a technique used to find the derivative dy/dx when y is defined implicitly as a function of x, rather than explicitly. How is the chain rule used in this example? The chain rule is applied to differentiate the composite function ln(x² + y²), requiring differentiation of both x² and y² terms and multiplying by dy/dx for the y terms. Why do we need to group terms with dy/dx on one side? Grouping terms with dy/dx on one side allows us to factor it out and solve algebraically for dy/dx, which is the derivative we want to find. ## Full Transcript — Download SRT & Markdown 00:00 Speaker A And I'm going to kind of work on this. We should be looking into taking this on both sides and looking into. So, if I'm going to take the derivative here with respect to x of y and d over dy. 00:19 Speaker A I'm sorry, DRDX is going to be ln of X squared plus Y squared. Okay, the derivative of Y with respect to X is DY DX. That's what we're trying to find. Cool. 00:35 Speaker A I'm sorry, dRdx is going to be ln of x squared plus y squared. Okay, the derivative of y with respect to x is dy dx. That's what we're trying to find. Cool. 00:48 Speaker A Right? And then times the derivative of the inside function. So, the derivative of the inside function, what's the derivative of X squared with respect to X? 00:59 Speaker A The derivative of ln of x squared plus y squared, so again, remember it's going to be basically, it's going to be 1 over x squared plus y squared. 01:12 Speaker A Think chain rule. Think chain rule. What's the derivative of Y squared? 2Y. What's the derivative of the inside function with respect to X? 01:22 Speaker A Right? And then times the derivative of the inside function. So, the derivative of the inside function, what's the derivative of x squared with respect to x? 01:40 Speaker A to X is DYDX. Okay? So, it's going to be 2X plus 2Y uh DYDX. 02:04 Speaker A 2x. What's the derivative of y squared with respect to x? Sorry. Let's do this over below. 02:18 Speaker A So what are we trying to solve for again? What are we take What are we trying to solve for? What have we What have they all have been trying to solve for? 02:27 Speaker A Think chain rule. Think chain rule. What's the derivative of y squared? 2y. What's the derivative of the inside function with respect to x? 02:39 Speaker A dy dx on the same side. Well guys, we kind of have a little bit of a problem here because we can't solve can't get the dy dx to the same side unless we distribute um this x squared minus y squared. 02:57 Speaker A dy over dx. Think chain rule. The derivative of d over dx of y squared is 2y, right? Bring down the two. But then you got to take the derivative of the inside function, which in this case is y. The derivative of y with respect to x is dy dx. Okay? So, it's going to be 2x plus 2y, uh, dy dx. 03:21 Speaker A equals 2x plus 2y dy dx. So again guys, to solve for dy dx you got to get them on the same side, right? 03:31 Speaker A So I have dy dx equals 2x plus 2y dy dx all over x squared plus y squared. 03:44 Speaker A So therefore I have dy dx times x squared plus y squared minus 2y dy dx = 2x. 03:57 Speaker A So what are we trying to solve for again? What are we take? What are we trying to solve for? What have we—what have they all been trying to solve for? 04:04 Speaker A Do what On the left side, what do they have in common? dy dx. So, I factor out of dy dx and I'm left with a x squared plus y squared 2y = 2x. Well, divide that on both sides, dy 04:31 Speaker A dy dx. We're trying to solve to find dy with respect to x. So when you guys apply, when you take the derivative, all right, you got to take the derivative of, you know, everything with respect to x. Then we got to group the 04:50 Speaker A And we could just write it or leave it um in that form. Yes. You multiplied by x squared plus y squared to get rid of the denominator. 05:02 Speaker A dy dx on the same side. Well, guys, we kind of have a little bit of a problem here because we can't solve—can't get the dy dx to the same side unless we distribute, um, this x squared minus y squared. 05:07 Speaker A Yes. Yeah, I wrote the x squared plus y squared on both sides to get rid of the denominator. 05:11 Speaker A So I multiply that on both sides. All right, and then what I would do is I would, uh, I would subtract a 2y, um. So anyways, let's look at it here. We have dy dx, um, times x squared plus y squared 05:19 Speaker A Okay? And then, could you show me the chain rule where the um d over dx Topics: implicit differentiation chain rule derivative dy/dx logarithmic differentiation calculus Brian McLogan math tutorial derivative of ln(x^2 + y^2) algebraic manipulation Study this video - [Flashcards from this video](https://sozai.app/tools/ai-flashcard-generator/?from=transcript&id=71501) - [Quiz on this video](https://sozai.app/tools/quiz-generator/?from=transcript&id=71501) Free, made by AI from this transcript. No signup. --- This is the markdown twin of https://sozai.app/transcript/implicit-differentiation-chain-rule/ — the same content, without the markup. Published by SozAI (https://sozai.app). Reuse and quotation are allowed with attribution and a link back. Machine-readable index: https://sozai.app/llms.txt · data API: https://sozai.app/api/