This lesson develops define successive derivatives and establish the notation and existence conditions for derivatives of finite order. The source order is retained, while the statements below make the required domains, regularity assumptions, and verification steps explicit. ## Higher-Order Derivatives and Notation :::definition[Second-Order Derivative] Let $f$ be differentiable on an interval. If $f'$ is differentiable, then the derivative of $f'$ is called the **second-order derivative** of $f$ and is denoted by $f''$. ::: :::definition[Nth-Order Derivative] Let $f,f',f'',\ldots,f^{(n-1)}$ be successively differentiable functions. The derivative of $f^{(n-1)}$ is called the **nth-order derivative** of $f$ and is denoted by $f^{(n)}$. ::: :::key-formula[Notation for Higher-Order Derivatives] $$\begin{equation*} f'(x)=\frac{dy}{dx}=y_1=Dy, \qquad f''(x)=\frac{d^2y}{dx^2}=y_2=D^2y, \qquad f^{(n)}(x)=\frac{d^ny}{dx^n}=y_n=D^ny. \end{equation*}$$ ::: :::key-formula[Successive Differentiation] $$\begin{equation*} \frac{d^ny}{dx^n} = \frac{d}{dx}\left(\frac{d^{n-1}y}{dx^{n-1}}\right) = \frac{d^2}{dx^2}\left(\frac{d^{n-2}y}{dx^{n-2}}\right). \end{equation*}$$ ::: :::definition[Existence of an Nth Derivative] The derivative $f^{(n)}(x)$ exists at $x$ only when $f^{(n-1)}$ exists in a neighbourhood of $x$ and is differentiable at $x$. ::: :::section[Method and Verification] State the domain, select the applicable differentiation rule, keep the order of differentiation explicit, and verify a general formula by checking low orders and differentiating once more. ::: :::solved-problem[Worked Verification] Let $f(x)=x^4-3x^2+2$. Determine $f'(x)$, $f''(x)$, $f^{(3)}(x)$, $f^{(4)}(x)$, and $f^{(5)}(x)$. ::: :::solution Differentiating successively, $$\begin{align*} f'(x)&=4x^3-6x,\\ f''(x)&=12x^2-6,\\ f^{(3)}(x)&=24x,\\ f^{(4)}(x)&=24,\\ f^{(5)}(x)&=0. \end{align*}$$ Therefore every derivative of order at least $5$ is identically zero. ::: :::remark[Common Error] Do not treat the superscript in $f^{(n)}$ as an exponent. Check the domain and the existence of every preceding derivative before applying an $n$th-derivative formula. ::: :::exercise[Practice Questions] 1. Find the first four derivatives of $f(x)=x^5-2x^3+x$. 2. Prove by induction that $D^n(e^{2x})=2^n e^{2x}$. 3. Show that $f(x)=|x|^{3/2}$ is differentiable at $0$ but does not possess a second derivative there. ::: :::answer[Answers and Solution Guidance] 1. $f'(x)=5x^4-6x^2+1$, $f''(x)=20x^3-12x$, $f^{(3)}(x)=60x^2-12$, and $f^{(4)}(x)=120x$. 2. The case $n=1$ holds. If $D^n(e^{2x})=2^n e^{2x}$, then differentiating once gives $D^{n+1}(e^{2x})=2^{n+1}e^{2x}$. 3. For $x\ne0$, $f'(x)=\frac32\operatorname{sgn}(x)|x|^{1/2}$ and $f'(0)=0$. The quotient $[f'(h)-f'(0)]/h$ is unbounded in magnitude as $h\to0$, so $f''(0)$ does not exist. ::: :::faq[Frequently Asked Questions] Q: What must be checked before using an nth-derivative formula? A: The function and every required preceding derivative must exist on the stated domain; additional restrictions such as nonzero denominators or positive logarithmic arguments must also hold. Q: How can a proposed nth derivative be verified? A: Check one or two initial orders and differentiate the proposed expression once to see whether it produces the formula with $n$ replaced by $n+1$. Q: Why is $f^{(n)}$ not the same as $f^n$? A: The notation $f^{(n)}$ denotes successive differentiation, whereas $f^n$ denotes an ordinary power of the function value. ::: :::review-history[Academic Review Notes] - Source mathematics and original concept order preserved. - Domains, hypotheses, exceptional cases, exercises, answers, FAQ pairs, and internal routes checked on 2026-07-22. ::: :::call-to-action[Continue Learning] subtitle: Complete the practice, verify the principal result, and review the available higher-order derivatives material. button: Review This Section|/real-analysis/functions-of-a-single-variable-ii/higher-order-derivatives button-ghost: Review the Chapter|/real-analysis/functions-of-a-single-variable-ii :::