Calculus Reference

Leibniz Notation for Derivatives

Leibniz notation identifies the quantity being differentiated, the variable it depends on, and the order of differentiation.

Notation at a Glance

Form	Use
$\frac{d}{dx}[f(x)]$	Instruction: differentiate $f(x)$ with respect to $x$.
$\frac{dy}{dx}=y'=f'(x)$	Result: the derivative function when $y=f(x)$.
$\left.\frac{dy}{dx}\right\|_{x=a}=f'(a)$	Value: the derivative evaluated at $x=a$.
$\frac{d^2y}{dx^2}$	Second derivative: the derivative of $\frac{dy}{dx}$.
$\frac{d^ny}{dx^n}$	$n$th derivative: differentiate $n$ times with respect to $x$.

From Differentiation Instruction to Result

$\frac{d}{dx}$ is an operator: it tells you to differentiate the expression that follows it. If $y=x^2+1$, applying the operator produces a derivative function.

Instruction

$$\frac{d}{dx}[y]=\frac{d}{dx}[x^2+1]$$

Result

$$\frac{dy}{dx}=y'=f'(x)=2x$$

The result still depends on $x$, so it gives a slope at every input. To get the slope at $x=2$, substitute $2$ after differentiating:

Derivative at One Input

$$\left.\frac{dy}{dx}\right|_{x=2}=f'(2)=2(2)=4$$

Differentials and Finite Changes

$\Delta x$ is a finite input change, and $\Delta y=f(x+\Delta x)-f(x)$ is the actual output change. In differential notation, choose an input differential $dx$ and define the corresponding output differential by $dy=f'(x)\,dx$. Thus $\frac{dy}{dx}=f'(x)$. For small input changes, $dy$ is the tangent-line estimate of $\Delta y$.

For $y=x^2$ at $x=3$, $\frac{dy}{dx}=6$. If $dx=0.1$, then $dy=6(0.1)=0.6$. The actual finite change is $\Delta y=(3.1)^2-3^2=0.61$.

Showing the Chain Rule

When $y$ depends on an intermediate variable $u$, and $u$ depends on $x$, the notation records each link. The product below is the chain rule; the matching $u$ marks the intermediate dependency.

Nested Function

$$u=3x+1,\qquad y=u^2$$

$$\frac{dy}{du}=2u,\qquad \frac{du}{dx}=3$$

$$\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}=(2u)(3)=6(3x+1)$$

The notation resembles fraction cancellation, but this product is justified by the chain rule.

When $y$ Is Not Isolated

If an equation relates $x$ and $y$, treat $y$ as a function of $x$ and apply $\frac{d}{dx}$ to both sides. The chain rule then supplies $\frac{dy}{dx}$ whenever a term contains $y$.

Circle

$$\frac{d}{dx}[x^2+y^2]=\frac{d}{dx}[25]$$

$$2x+2y\frac{dy}{dx}=0$$

$$\frac{dy}{dx}=-\frac{x}{y},\qquad y\ne0$$

In $\frac{d}{dx}[y^2]=2y\frac{dy}{dx}$, the factor $\frac{dy}{dx}$ appears because $y$ changes as $x$ changes. See implicit differentiation for further examples.

Second and Higher Derivatives

Differentiate the first derivative again to get the second derivative. For $y=x^3$:

Repeated Differentiation

$$\frac{dy}{dx}=3x^2$$

$$\frac{d^2y}{dx^2}=\frac{d}{dx}\left[\frac{dy}{dx}\right]=6x$$

$$\frac{d^3y}{dx^3}=6$$

The superscript $2$ in $\frac{d^2y}{dx^2}$ marks the second derivative; it does not square the first derivative. Here, $\left(\frac{dy}{dx}\right)^2=9x^4$, which differs from $\frac{d^2y}{dx^2}=6x$.