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For a fixed unit direction $v$, the multivariable Taylor expansion gives the displayed leading error. Equal directional derivatives in two directions do not imply equal approximation errors: the linear changes depend on the gradient, while the quadratic errors depend on the Hessian matrix. For example, $f(x,y)=x+y+2x^2+y^2$ at the origin has equal derivatives in the two coordinate directions but errors $2s^2$ and $s^2$.

The first-order multivariable Taylor expansion isTaking $(x_0,y_0)=(2,0)$ gives the linear approximationTo assess its significant figures, the quadratic Taylor polynomial supplies the correction $h^2+4hk+2k^2=-0.0001$ for $h=0.03$, $k=-0.01$. Direct evaluation gives $2.03^2{e}^{-0.01}=4.0798963599\dots$. Thus the estimate is correct to four significant figures: both round to $4.080$, whereas to five significant figures the exact value is $4.0799$ and the approximation is $4.0800$. The unexpectedly small quadratic correction results partly from cancellation of its mixed and unmixed terms.

For $f$ with continuous second partial derivatives near $(a,b)$, its multivariable Taylor expansion iswhere all displayed derivatives are evaluated at $(a,b)$. Continuity of the second derivatives justifies equality of the mixed partials and the stated remainder. A cubic-order remainder requires additional regularity.

Put $h=x-1/\sqrt{2}$ and $k=y-1/(2\sqrt{2})$. At the expansion point the exponent is one and the gradient is $(\sqrt{2}e,\sqrt{2}e)$. The first-order multivariable Taylor expansion isThe directional error of a first-order Taylor approximation is determined by the omitted terms, not just the equal first partial derivatives. The Hessian matrix at the expansion point is $e\left(\begin{array}{cc}4& 2\\ 2& 2\end{array}\right)$, soFor equal small steps along the coordinate axes, the leading error is $2e{h}^2$ along $x$ and $e{k}^2$ along $y$. The error grows faster along the $x$ axis, although the first-order change in the function is equal along the two axes.