The expression fdiff(sin(x), x=1) computes a numerical approximation to sin'(1) = cos(1) without using a formula for diff(sin(x),x). It does this by evaluating f at two points near $x=1$. More generally, the command fdiff(f(x,y,z),[x,x,y],[x=a,y=b,z=c] computes a numerical approximation for the partial derivative diff(f(x,y,z), x,x,y) evaluated at $\left\{x=a\,y=b\,z=c\right\}$. The differentiation is not performed symbolically, and the computation of this approximation assumes only that f can be evaluated numerically to high precision near the point $x=a,y=b,z=c$.

The fdiff command is useful when a closed form formula is not known for $f\'\left(x\right)$, thus for instance, permitting the plotting of expressions containing such derivatives. The accuracy of the approximation is determined by the global variable Digits and the optional argument workprec which is described in the Computational Approach section following. The algorithm used will normally, but not necessarily, output a result accurate to Digits precision.

The first argument passed to fdiff is the expression being differentiated. The second argument can be a set or a list of the differentiation variables. It specifies the order of the derivative.  The third argument can be set or a list with equations containing the differentiation variables on the left-hand sides and the evaluation points on the right-hand sides.  It specifies the point at which the derivative is computed.  For example,

Note: The order of the derivative is always equal to the number of elements in the second argument, and the equations in the third argument should assure that all the names in the expression being differentiated get a value, and the evaluation points ($a,b$ in the above) should be numerical constants so that numerical differentiation can be performed. If the numerical derivative cannot be computed, fdiff returns unevaluated.

The right-hand side in the correspondence above can also be entered as an argument to evalf, for example,

in which case it is (internally) transformed into the corresponding call to fdiff, then numerically evaluated. The argument to evalf can also involve Eval instead of eval and Diff instead of diff.

Alternatively, the following shorter syntax can be used:

For simplicity, an alternative syntax is allowed: the second argument, $\left[x\,y\right]$ in fdiff( f(x,y), [x,y], {x=a, y=b}) can be omitted, in which case the third argument, $\left\{x=a\,y=b\right\}$ indicates, simultaneously, the differentiations to be performed and the evaluation points. For example, these two calls result in the same output

Note: In the right-hand side, every equation entering the second argument implies a differentiation.

When the expression being differentiated depends on one variable only, or the computation is a first derivative, the second or third arguments can also be given without enclosing them as sets or lists, for example,

The expression being differentiated can also be a procedure or operator, which is a function of one or more inputs. This is useful when a function cannot be expressed as a formula and also when a program would be a more efficient representation of a function than a formula. In this case, the differentiation variables are specified in the second argument as a list with their numeric positions (see D), and the third argument to fdiff should be a list with the constant values corresponding to each (dummy) variable in the procedure. So, for instance, the correspondence is as in

The method used is a fixed point method.  That is, for a given order of derivative, the function f is evaluated at a fixed number of points near the evaluation point $x=a,y=b,\dots$ at sufficient precision so that all digits in the output are accurate. However, if the function is changing rapidly near the evaluation point, the result may be inaccurate. The optional argument workprec=n where $n$ is a numerical constant > 1 means that the computation will be computed at a working precision of n*Digits precision and the result rounded to Digits precision. The default value for workprec is 1.

The following describes how the differentiation is performed numerically to compute D(f)(a). The Taylor series for $f\left(x+h\right)$ and $f\left(x-h\right)$ about $h=0$ are

To estimate $f\'\left(x\right)$, solve either of the above for $f\'\left(x\right)$, giving a one-sided difference formula, for example, $f\text{'}\left(x\right)=\frac{f\left(x+h\right)-f\left(x\right)}{h}-\frac{h}{2}\cdot f\text{'}\text{'}\left(x\right)+\dots$ with error term $\frac{h}{2}f\text{'}\text{'}\left(x\right)+\mathrm{...}=\mathrm{O}\left(h\right)$ and requiring two function evaluations. Instead, fdiff uses a symmetric difference, taking (1) - (2) to obtain

which also requires two function evaluations but has an error term $\frac{h^2}{6}f\text{'}\text{'}\text{'}\left(x\right)+\mathrm{...}$, which is $\mathrm{O}\left(h^2\right)$.

Let $n=\mathrm{Digits}$ and $e={10}^{-n}$, that is, $e$ is a machine epsilon.

To compute $f\'\left(x\right)$ accurate to $n$ digits, take $h={10}^{-\frac{n}{2}}=\sqrt{e}$ in ( * ). Thus the $\mathrm{error}=\frac{h^2}{6}f\text{'}\text{'}\text{'}\left(x\right)+\mathrm{...}=\mathrm{O}\left(h^2\right)=\mathrm{O}\left({10}^{-n}\right)=\mathrm{O}\left(e\right)$. Note that $\frac{f\left(x+\frac{h}{2}\right)-f\left(x-\frac{h}{2}\right)}{h}$ loses $\frac{n}{2}$ digits of precision. Hence, to obtain the difference with $n$ digits of precision, fdiff must compute $f\left(x+\frac{h}{2}\right)$ and $f\left(x-\frac{h}{2}\right)$ to at least $\frac{3n}{2}$ digits of precision. The actual values for the precision used by fdiff to compute the linear combination of the $f$'s allow for two guard digits.

For the second derivative, $f\text{'}\text{'}\left(x\right)$, solving (1) + (2) for $f\text{'}\text{'}\left(x\right)$ one has

with error term $\mathrm{O}\left(h^2\right)$ but requiring three function calls. Setting $h={10}^{-\frac{n}{2}}$, the $\mathrm{error}=\frac{h^2}{12}f\text{'}\text{'}\text{'}\text{'}\left(x\right)+\mathrm{...}=\mathrm{O}\left(h^2\right)={10}^{-n}$.

For higher order derivatives one may locate the evaluation points symmetrically and equally spaced about $x$, namely, for even order derivatives, at $\mathrm{...},x-2h,x-h,x,x+h,x+2h,\mathrm{...}$, and for odd order derivatives, at $\mathrm{...},x-\frac{3h}{2},x-\frac{h}{2},x+\frac{h}{2},x+\frac{3h}{2},\mathrm{...}$ leading to an error term $\mathrm{O}\left(h^2\right)$.

$\mathrm{fdiff}\left(\cos \left(x\right)\,x=1\right)$

$\mathrm{-0.8414709848}$

$\mathrm{diff}\left(\cos \left(x\right)\,x\right)$

$-\sin \left(x\right)$

$\mathrm{evalf}\left(\mathrm{eval}\left(\,x=1\right)\right)$

$\mathrm{-0.8414709848}$

$\mathrm{fdiff}\left(\mathrm{BesselJ}\left(a\,\frac{1}{2}\right)\,a=\frac{1}{3}\right)=\mathrm{eval}\left(\mathrm{diff}\left(\mathrm{BesselJ}\left(a\,\frac{1}{2}\right)\,a\right)\,a=\frac{1}{3}\right)$

$\mathrm{-0.8196217810}=\genfrac{}{}{0ex}{}{\left(\frac{\ⅆ}{\ⅆa}\mathrm{BesselJ}\left(a\,\frac{1}{2}\right)\right)}{\phantom{\left\{a=\frac{1}{3}\right\}}}|\genfrac{}{}{0ex}{}{\phantom{\left(\frac{\ⅆ}{\ⅆa}\mathrm{BesselJ}\left(a\,\frac{1}{2}\right)\right)}}{\left\{a=\frac{1}{3}\right\}}$

$z≔\mathrm{eval}\left(\mathrm{diff}\left(\mathrm{BesselJ}\left(a\,\frac{1}{2}\right)\,a\right)\,a=\frac{1}{3}\right)$

$z≔\genfrac{}{}{0ex}{}{\left(\frac{\ⅆ}{\ⅆa}\mathrm{BesselJ}\left(a\,\frac{1}{2}\right)\right)}{\phantom{\left\{a=\frac{1}{3}\right\}}}|\genfrac{}{}{0ex}{}{\phantom{\left(\frac{\ⅆ}{\ⅆa}\mathrm{BesselJ}\left(a\,\frac{1}{2}\right)\right)}}{\left\{a=\frac{1}{3}\right\}}$

$\mathrm{evalf}\left(z\right)$

$\mathrm{-0.8196217810}$

$\mathrm{evalf}\left(\mathrm{D}\left[1\right]\left(\mathrm{BesselJ}\right)\left(\frac{1}{3}\,\frac{1}{2}\right)\right)$

$\mathrm{-0.8196217810}$

$z≔\mathrm{Eval}\left(\mathrm{Diff}\left(\mathrm{BesselJ}\left(a\,\frac{1}{2}\right)\,a\right)\,a=\frac{1}{3}\right)$

$z≔\genfrac{}{}{0ex}{}{\left(\frac{\ⅆ}{\ⅆa}\mathrm{BesselJ}\left(a\,\frac{1}{2}\right)\right)}{\phantom{a=\frac{1}{3}}}|\genfrac{}{}{0ex}{}{\phantom{\left(\frac{\ⅆ}{\ⅆa}\mathrm{BesselJ}\left(a\,\frac{1}{2}\right)\right)}}{a=\frac{1}{3}}$

$\mathrm{evalf}\left(z\right)$

$\mathrm{-0.8196217810}$

$\mathrm{eval}\left(\mathrm{diff}\left(\mathrm{LegendreP}\left(a\,b\,x\right)\,a\,a\,b\,b\right)\,\left[a=\frac{1}{2}\,b=\frac{3}{2}\,x=\frac{5}{7}\right]\right)=\mathrm{fdiff}\left(\mathrm{LegendreP}\left(a\,b\,x\right)\,\left[a\,a\,b\,b\right]\,\left[a=\frac{1}{2}\,b=\frac{3}{2}\,x=\frac{7}{5}\right]\right)$

$\genfrac{}{}{0ex}{}{\left(\frac{{\partial }^4}{\partial a^2\partial b^2}\mathrm{LegendreP}\left(a\,b\,\frac{5}{7}\right)\right)}{\phantom{\left\{a=\frac{1}{2}\,b=\frac{3}{2}\right\}}}|\genfrac{}{}{0ex}{}{\phantom{\left(\frac{{\partial }^4}{\partial a^2\partial b^2}\mathrm{LegendreP}\left(a\,b\,\frac{5}{7}\right)\right)}}{\left\{a=\frac{1}{2}\,b=\frac{3}{2}\right\}}=\mathrm{-3.553866479}$

$\mathrm{D}\left[1,1,2,2\right]\left(\mathrm{LegendreP}\right)\left(\frac{1}{2}\,\frac{3}{2}\,\frac{7}{5}\right)=\mathrm{fdiff}\left(\mathrm{LegendreP}\,\left[1\,1\,2\,2\right]\,\left[\frac{1}{2}\,\frac{3}{2}\,\frac{7}{5}\right]\right)$

${\mathrm{D}}_{1,1,2,2}\left(\mathrm{LegendreP}\right)\left(\frac{1}{2}\,\frac{3}{2}\,\frac{7}{5}\right)=\mathrm{-3.553866479}$

g := proc(x,a,n::integer) local y;
        if n=0 then x else y := g(x,a,n-1); a*y*(1-y) end if;
     end proc:

$g\left(x\,a\,3\right)$

$a^{3}x\left(1-x\right)\left(1-ax\left(1-x\right)\right)\left(1-a^{2}x\left(1-x\right)\left(1-ax\left(1-x\right)\right)\right)$

$\mathrm{degree}\left(g\left(x\,a\,3\right)\,x\right),\mathrm{degree}\left(g\left(x\,a\,3\right)\,a\right)$

$8,7$

$\mathrm{degree}\left(g\left(x\,a\,4\right)\,x\right),\mathrm{degree}\left(g\left(x\,a\,4\right)\,a\right)$

$16,15$

$\mathrm{degree}\left(g\left(x\,a\,5\right)\,x\right),\mathrm{degree}\left(g\left(x\,a\,5\right)\,a\right)$

$32,31$

$\mathrm{D}\left[2\right]\left(g\right)\left(\frac{1}{3}\,\frac{1}{2}\,5\right)=\mathrm{fdiff}\left(g\,\left[1\,2\right]\,\left[\frac{1}{3}\,\frac{1}{2}\,5\right]\right)$

${\mathrm{D}}_2\left(g\right)\left(\frac{1}{3}\,\frac{1}{2}\,5\right)=0.05826998816$

$f≔9000x^5+1$

$f≔9000x^5+1$

$a≔\frac{500000000}{3}$

$a≔\frac{500000000}{3}$

$\mathrm{answer}≔\mathrm{evalf}\left[20\right]\left(\mathrm{eval}\left(\mathrm{diff}\left(f\,x\right)\,x=a\right)\right)$

$\mathrm{answer}≔3.4722222222222222222\times {10}^{37}$

$\mathrm{fdiff}\left(f\,x=a\right)$

$3.47200000\times {10}^{37}$

$\mathrm{fdiff}\left(f\,x=a\,\mathrm{workprec}=1.1\right)$

$3.472220000\times {10}^{37}$

$\mathrm{fdiff}\left(f\,x=a\,\mathrm{workprec}=1.3\right)$

$3.472222200\times {10}^{37}$

$\mathrm{fdiff}\left(f\,x=a\,\mathrm{workprec}=1.5\right)$

$3.472222222\times {10}^{37}$

$a≔'a'\:$

$f≔\mathrm{diff}\left(\mathrm{BesselK}\left(a\,\frac{3}{5}\right)\,a\right)\:$

$f=\mathrm{convert}\left(\mathrm{series}\left(f\,a\right)\,\mathrm{compose}\,\mathrm{diff}\,\mathrm{polynom}\right)$

$\frac{\ⅆ}{\ⅆa}\mathrm{BesselK}\left(a\,\frac{3}{5}\right)=\genfrac{}{}{0ex}{}{\left(\frac{\ⅆ}{\ⅆ\mathrm{t1}}\mathrm{BesselK}\left(\mathrm{t1}\,\frac{3}{5}\right)\right)}{\phantom{\left\{\mathrm{t1}=0\right\}}}|\genfrac{}{}{0ex}{}{\phantom{\left(\frac{\ⅆ}{\ⅆ\mathrm{t1}}\mathrm{BesselK}\left(\mathrm{t1}\,\frac{3}{5}\right)\right)}}{\left\{\mathrm{t1}=0\right\}}+\left(\genfrac{}{}{0ex}{}{\left(\frac{{\ⅆ}^2}{\ⅆ{\mathrm{t1}}^2}\mathrm{BesselK}\left(\mathrm{t1}\,\frac{3}{5}\right)\right)}{\phantom{\left\{\mathrm{t1}=0\right\}}}|\genfrac{}{}{0ex}{}{\phantom{\left(\frac{{\ⅆ}^2}{\ⅆ{\mathrm{t1}}^2}\mathrm{BesselK}\left(\mathrm{t1}\,\frac{3}{5}\right)\right)}}{\left\{\mathrm{t1}=0\right\}}\right)a+\frac{\left(\genfrac{}{}{0ex}{}{\left(\frac{{\ⅆ}^3}{\ⅆ{\mathrm{t1}}^3}\mathrm{BesselK}\left(\mathrm{t1}\,\frac{3}{5}\right)\right)}{\phantom{\left\{\mathrm{t1}=0\right\}}}|\genfrac{}{}{0ex}{}{\phantom{\left(\frac{{\ⅆ}^3}{\ⅆ{\mathrm{t1}}^3}\mathrm{BesselK}\left(\mathrm{t1}\,\frac{3}{5}\right)\right)}}{\left\{\mathrm{t1}=0\right\}}\right)a^2}{2}+\frac{\left(\genfrac{}{}{0ex}{}{\left(\frac{{\ⅆ}^4}{\ⅆ{\mathrm{t1}}^4}\mathrm{BesselK}\left(\mathrm{t1}\,\frac{3}{5}\right)\right)}{\phantom{\left\{\mathrm{t1}=0\right\}}}|\genfrac{}{}{0ex}{}{\phantom{\left(\frac{{\ⅆ}^4}{\ⅆ{\mathrm{t1}}^4}\mathrm{BesselK}\left(\mathrm{t1}\,\frac{3}{5}\right)\right)}}{\left\{\mathrm{t1}=0\right\}}\right)a^3}{6}+\frac{\left(\genfrac{}{}{0ex}{}{\left(\frac{{\ⅆ}^5}{\ⅆ{\mathrm{t1}}^5}\mathrm{BesselK}\left(\mathrm{t1}\,\frac{3}{5}\right)\right)}{\phantom{\left\{\mathrm{t1}=0\right\}}}|\genfrac{}{}{0ex}{}{\phantom{\left(\frac{{\ⅆ}^5}{\ⅆ{\mathrm{t1}}^5}\mathrm{BesselK}\left(\mathrm{t1}\,\frac{3}{5}\right)\right)}}{\left\{\mathrm{t1}=0\right\}}\right)a^4}{24}+\frac{\left(\genfrac{}{}{0ex}{}{\left(\frac{{\ⅆ}^6}{\ⅆ{\mathrm{t1}}^6}\mathrm{BesselK}\left(\mathrm{t1}\,\frac{3}{5}\right)\right)}{\phantom{\left\{\mathrm{t1}=0\right\}}}|\genfrac{}{}{0ex}{}{\phantom{\left(\frac{{\ⅆ}^6}{\ⅆ{\mathrm{t1}}^6}\mathrm{BesselK}\left(\mathrm{t1}\,\frac{3}{5}\right)\right)}}{\left\{\mathrm{t1}=0\right\}}\right)a^5}{120}$

$\mathrm{plots}\left[\mathrm{plotcompare}\right]\left(\right)$