Q.Let $V(x,y,z)=xy+yz+zx,\ x,y,z\in\mathbb R$. Find the differential $dV$.

A nonlinear function is generally hard to evaluate exactly, but near any one point its graph looks almost like a straight line -- the tangent line at that point. This is the idea behind linear approximation.

Definition (Linear Approximation). Let $f:(a,b)\to\mathbb R$ be differentiable and $x_0\in(a,b)$. The linear approximation $L$ of $f$ at $x_0$ is

This is exactly the equation of the tangent line to $y=f(x)$ at $(x_0,f(x_0))$. Because $f$ is differentiable, $f(x_0+\Delta x)\approx L(x_0+\Delta x) = f(x_0)+f'(x_0)\Delta x$ when $\Delta x$ is small, and the error $f(x)-L(x)$ shrinks to $0$ faster than $x\to x_0$ (it is $o(x-x_0)$, a consequence of the definition of the derivative as a limit).

The differential. Writing $\Delta x = dx$, the tangent-line increment is called the differential of $f$:

Geometrically, $\Delta f = f(x+dx)-f(x)$ is the actual rise along the curve, while $df=f'(x)dx$ is the rise along the tangent line; for small $dx$, $\Delta f\approx df$, but the two are generally not equal (only exactly equal when $f$ is itself linear, $f(x)=mx+c$). $df$ is a function of two independent quantities, $x$ and $dx$ -- not of $x$ alone the way the derivative is.

Differentials of standard functions follow directly from the derivative rules, e.g. $d(x^n)=nx^{n-1}dx$, $d(\sin x)=\cos x\,dx$, $d(e^x)=e^x dx$, $d(\log x)=\dfrac1x dx$. The algebraic properties of differentials mirror differentiation exactly:

• $d(c)=0$ for a constant $c$; $\ d(x)=dx$.
• $d(cf) = c\,df$.
• $d(f\pm g) = df\pm dg$.
• Product rule: $d(fg) = f\,dg + g\,df$.
• Quotient rule: $d\!\left(\dfrac fg\right) = \dfrac{g\,df-f\,dg}{g^2}$, $g\ne0$.
• Chain rule: if $h=f\circ g$, $dh = f'(g(x))\,g'(x)\,dx$. …