Second-Order Composite Functions is a chronological lesson in higher-order differentiation. The source manuscript supplies this topic as part of Methods and Applications of Higher-Order Differentiation, and the presentation below preserves the mathematical intention while making hypotheses explicit. Let $u=g(x)$ and $y=f(u)$. Then $y=f(g(x))$ is a composite function and $u$ is the intermediate variable. :::theorem[Second-Order Chain Rule] Let $u=g(x)$ and $y=f(u)$ be twice differentiable functions. Then $$\begin{equation*} \frac{d^2y}{dx^2} = \frac{d^2u}{dx^2}\frac{dy}{du} + \left(\frac{du}{dx}\right)^2 \frac{d^2y}{du^2}. \end{equation*}$$ ::: :::proof Given that $y=f(u)$ and $u=g(x)$. By the first-order chain rule, $$\begin{equation*} \frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}. \end{equation*}$$ Differentiating with respect to $x$, $$\begin{equation*} \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{du}\right)\frac{du}{dx} + \frac{dy}{du}\frac{d^2u}{dx^2}. \end{equation*}$$ Again by the chain rule, $$\begin{equation*} \frac{d}{dx}\left(\frac{dy}{du}\right) = \frac{d^2y}{du^2}\frac{du}{dx}. \end{equation*}$$ Substituting gives the stated formula. $\square$ ::: :::example[Transformation with a Sine Variable] Let $u=\sin x$ and $y=u^2$. Then $$\begin{equation*} \frac{dy}{du}=2u,\qquad \frac{d^2y}{du^2}=2,\qquad \frac{du}{dx}=\cos x,\qquad \frac{d^2u}{dx^2}=-\sin x. \end{equation*}$$ Therefore $$\begin{equation*} \frac{d^2y}{dx^2}=(-\sin x)(2\sin x)+2\cos^2x=2\cos 2x. \end{equation*}$$ ::: --- :::exercise[Practice Questions] 1. State the principal formula or definition used in Second-Order Composite Functions. 2. Apply the main method of this lesson to one simple example of your own choice. 3. Explain one domain, branch, or nonzero-denominator condition that must be checked before using the result. ::: :::answer[Answers and Guidance] 1. Write the formula with all hypotheses before substituting values. 2. Differentiate or simplify step by step, and verify the result by checking a low-order case. 3. The condition must be inherited from the original function or from a division step in the derivation; it must not be added after the computation. ::: :::faq[Frequently Asked Questions] Q: What is the main purpose of this lesson? A: The purpose is to use second-order composite functions in a controlled way, with all variables, orders, and restrictions stated before calculation. Q: What is the most common error in this topic? A: The most common error is to apply a formula without checking the differentiability, domain, branch, or nonzero-denominator condition required by the formula. Q: How should I verify an answer? A: Differentiate the obtained expression in a low-order case, compare it with the stated formula, and confirm that the excluded values have not been lost. ::: :::call-to-action[Continue Learning] subtitle: Revise the formulae, complete the practice questions, and continue to the next lesson in higher-order differentiation. button: Next Lesson|/real-analysis/functions-of-a-single-variable-ii/higher-order-derivatives/derivatives-of-inverse-functions button-ghost: Review Higher Order Derivatives|/real-analysis/functions-of-a-single-variable-ii/higher-order-derivatives :::