Calculus

This tutorial climbs from algebra into calculus. We differentiate, integrate, expand functions as power series, take limits, evaluate sums, and finish with numerical root-finding and optimisation. All of it is built on the same expression-rewriting engine you met in the algebra tutorial — calculus in Mathilda is mostly a large, well-organised collection of rewrite rules.

Every transcript below was produced by the actual Mathilda binary. Type the In[...] lines yourself and you will see the same Out[...]. Mathilda's output form is sometimes arranged differently from a textbook (terms may be reordered, E^x is written instead of Exp[x], and indefinite integrals may differ by a constant), but it is always mathematically correct.

Differentiation with D

D[expr, x] differentiates expr with respect to x. Mathilda knows the standard rules of calculus and applies them recursively, so you can throw fairly involved expressions at it. Start with the power rule, which works for symbolic exponents too:

In[1]:= D[x^5, x]
Out[1]= 5 x^4

In[2]:= D[x^n, x]
Out[2]= n x^(-1 + n)

The product rule appears automatically when you differentiate a product, and the chain rule when a function's argument is itself a function of x:

In[1]:= D[x Sin[x], x]
Out[1]= x Cos[x] + Sin[x]

In[2]:= D[Sin[x^2], x]
Out[2]= 2 x Cos[x^2]

In[3]:= D[Exp[3 x], x]
Out[3]= 3 E^(3 x)

In In[1] each factor is differentiated in turn — x contributes Sin[x] and Sin[x] contributes x Cos[x] — and the results are summed. In In[2] the outer derivative Cos[x^2] is multiplied by the inner derivative 2 x; the same logic produces the factor of 3 in In[3].

The elementary functions all carry their derivatives:

In[1]:= D[Log[x], x]
Out[1]= 1/x

In[2]:= D[Tan[x], x]
Out[2]= Sec[x]^2

In[3]:= D[ArcTan[x], x]
Out[3]= 1/(1 + x^2)

Higher-order derivatives

To differentiate more than once, use the {x, n} form, which computes the n-th derivative with respect to x. Here is the fourth derivative of sine, which — after four quarter-turns through cosine and back — returns to sine:

In[1]:= D[Sin[x], {x, 4}]
Out[1]= Sin[x]

Partial derivatives

When an expression contains several variables, D[expr, x] differentiates with respect to x and treats the others as constants:

In[1]:= D[x^2 y^3, x]
Out[1]= 2 x y^3

In[2]:= D[x^2 y^3, y]
Out[2]= 3 x^2 y^2

In In[1] the factor y^3 rides along unchanged; in In[2] it is x^2 that is held constant. This is exactly the partial-derivative convention.

Derivatives of unknown functions

Differentiating an abstract function — one with no definition — produces Derivative notation rather than a concrete answer, which is how the chain and product rules express themselves in general:

In[1]:= D[f[g[x]], x]
Out[1]= Derivative[1][g][x] Derivative[1][f][g[x]]

In[2]:= D[f[x], {x, 2}]
Out[2]= Derivative[2][f][x]

Derivative[1][f] is "the first derivative of f", and Derivative[2][f] the second. In[1] is precisely the chain rule f'(g(x)) · g'(x) written in Mathilda's notation.

Total derivatives with Dt

D computes partial derivatives. Its cousin Dt computes total differentials, treating every symbol as potentially depending on the differentiation variable:

In[1]:= Dt[x y]
Out[1]= Dt[x] y + x Dt[y]

In[2]:= Dt[x^2, x]
Out[2]= 2 x

In[1] gives the total differential of a product — y dx + x dy in classical notation. In[2] takes the total derivative with respect to x; with no other dependencies declared, it agrees with D.

Integration with Integrate

Integrate[expr, x] computes an indefinite integral (an antiderivative) with respect to x. Mathilda omits the constant of integration, as is conventional in symbolic systems — so two correct answers may differ by a constant. Integration is genuinely harder than differentiation (there is no single mechanical rule), so Mathilda runs a cascade of methods.

Polynomials and the basic functions are the easy cases:

In[1]:= Integrate[x^4, x]
Out[1]= 1/5 x^5

In[2]:= Integrate[1/x, x]
Out[2]= Log[x]

In[3]:= Integrate[Cos[x], x]
Out[3]= Sin[x]

In[4]:= Integrate[Exp[x], x]
Out[4]= E^x

Two more worth committing to memory: the reciprocal of x^2 + 1 integrates to an inverse tangent, and 1/(x^2 - 1) to an inverse hyperbolic tangent:

In[1]:= Integrate[1/(x^2 + 1), x]
Out[1]= ArcTan[x]

In[2]:= Integrate[1/(x^2 - 1), x]
Out[2]= -ArcTanh[x]

For integrands that need integration by parts, Mathilda applies the technique automatically. Two classic cases are x e^x and x^2 log x:

In[1]:= Integrate[x Exp[x], x]
Out[1]= -E^x + x E^x

In[2]:= Integrate[x^2 Log[x], x]
Out[2]= 1/9 (-x^3 + 3 x^3 Log[x])

You can always check an integral by differentiating it — antidifferentiation and differentiation are inverses, so the derivative of the answer must recover the integrand:

In[1]:= D[-E^x + x E^x, x]
Out[1]= x E^x

The derivative of In[1] of the previous block returns x E^x, confirming the integration by parts.

A rational function is integrated by first splitting it into partial fractions (the Apart of the algebra tutorial) and integrating each piece. Mathilda does this for you:

In[1]:= Integrate[(2 x + 3)/((x + 1)(x + 2)), x]
Out[1]= Log[2 + 3 x + x^2]

Recall that Apart[(2 x + 3)/((x + 1)(x + 2))] is 1/(x + 1) + 1/(x + 2); integrating those two pieces gives Log[x + 1] + Log[x + 2], which combines into the single logarithm Mathilda reports (its argument 2 + 3x + x^2 is just (x + 1)(x + 2) expanded).

Indefinite only

Integrate in this build computes indefinite integrals — it does not accept integration bounds. To evaluate a definite integral, find the antiderivative with Integrate and apply the limits yourself.

Series expansions

Series[f, {x, a, n}] expands f as a power series in x around the point a, keeping terms up to order n. The trailing O[x]^... records the order of the first dropped term, so you always know how far the approximation is trusted.

The exponential is the cleanest example — every coefficient is 1/k! — while cosine keeps only even powers and Log[1 + x] has the alternating harmonic coefficients:

In[1]:= Series[Exp[x], {x, 0, 5}]
Out[1]= 1 + x + 1/2 x^2 + 1/6 x^3 + 1/24 x^4 + 1/120 x^5 + O[x]^6

In[2]:= Series[Cos[x], {x, 0, 6}]
Out[2]= 1 - 1/2 x^2 + 1/24 x^4 - 1/720 x^6 + O[x]^7

In[3]:= Series[Log[1 + x], {x, 0, 4}]
Out[3]= x - 1/2 x^2 + 1/3 x^3 - 1/4 x^4 + O[x]^5

The geometric series and the arctangent series are textbook results, and Mathilda even expands functions like Tan whose coefficients have no simple closed form:

In[1]:= Series[1/(1 - x), {x, 0, 5}]
Out[1]= 1 + x + x^2 + x^3 + x^4 + x^5 + O[x]^6

In[2]:= Series[ArcTan[x], {x, 0, 5}]
Out[2]= x - 1/3 x^3 + 1/5 x^5 + O[x]^6

In[3]:= Series[Tan[x], {x, 0, 5}]
Out[3]= x + 1/3 x^3 + 2/15 x^5 + O[x]^6

You can expand around a point other than the origin. The series is then written in powers of (x - a):

In[1]:= Series[Exp[x], {x, 1, 2}]
Out[1]= E + E (x - 1) + 1/2 E (x - 1)^2 + O[x - 1]^3

Every coefficient is a power of E, because all the derivatives of e^x at x = 1 equal e. To turn a series back into an ordinary polynomial — dropping the O[x] bookkeeping term — apply Normal:

In[1]:= Normal[Series[Exp[x], {x, 0, 3}]]
Out[1]= 1 + x + 1/2 x^2 + 1/6 x^3

The result is now a plain expression you can differentiate, integrate, or evaluate like any other.

Limits

Limit[expr, x -> a] evaluates the limiting value of expr as x approaches a. Its real value is computing indeterminate forms — expressions that are 0/0 or ∞/∞ on direct substitution. The most famous is Sin[x]/x at the origin:

In[1]:= Limit[Sin[x]/x, x -> 0]
Out[1]= 1

In[2]:= Limit[(1 - Cos[x])/x^2, x -> 0]
Out[2]= 1/2

In[3]:= Limit[(Exp[x] - 1)/x, x -> 0]
Out[3]= 1

All three are 0/0 if you just substitute, yet each has a finite limit that Limit finds. Limits at infinity are equally at home — use the symbol Infinity as the target:

In[1]:= Limit[(2 x^2 + 1)/(x^2 + x), x -> Infinity]
Out[1]= 2

In[2]:= Limit[(1 + 1/x)^x, x -> Infinity]
Out[2]= E

In[1] compares the growth rates of numerator and denominator: as x grows, the lower-order terms become negligible and the ratio tends to that of the leading coefficients, 2. In[2] is the celebrated limit that defines Euler's number.

When a limit is genuinely infinite, Mathilda says so. A real-signed blow-up gives Infinity; a blow-up with no single direction gives ComplexInfinity:

In[1]:= Limit[Log[x], x -> 0]
Out[1]= -Infinity

In[2]:= Limit[1/x, x -> 0]
Out[2]= ComplexInfinity

Log[x] decreases without bound as x -> 0, so In[1] is -Infinity. 1/x runs to +∞ from the right but -∞ from the left, so it has no ordinary two-sided limit — Mathilda reports ComplexInfinity to signal exactly that.

Sums

Sum[expr, {k, lo, hi}] adds up expr as the index k runs from lo to hi. With concrete numeric bounds you simply get a number:

In[1]:= Sum[k^2, {k, 1, 10}]
Out[1]= 385

The real power, though, is a symbolic upper bound — Mathilda returns a closed form. The sums of the first n integers, squares, and cubes all have classic formulas:

In[1]:= Sum[k, {k, 1, n}]
Out[1]= 1/2 n (1 + n)

In[2]:= Sum[k^2, {k, 1, n}]
Out[2]= 1/6 n (1 + n) (1 + 2 n)

In[3]:= Sum[k^3, {k, 1, n}]
Out[3]= 1/4 n^2 (1 + n)^2

The last is the lovely identity that the sum of the first n cubes equals the square of the sum of the first n integers. Geometric sums work too, even with a symbolic base:

In[1]:= Sum[2^k, {k, 0, n}]
Out[1]= -1 + 2^(1 + n)

In[2]:= Sum[x^k, {k, 0, n}]
Out[2]= -1/(-1 + x) + x^(1 + n)/(-1 + x)

In[1] is the familiar 2^(n+1) - 1. Rearranged, In[2] is (x^(n+1) - 1)/(x - 1) — the closed form for a finite geometric series, derived entirely symbolically.

Finite sums

Sum in this build evaluates finite sums: the bounds (or a symbolic upper bound like n) must be supplied. Infinite sums with an Infinity bound are left unevaluated.

Numerical calculus

Not every equation has a closed-form solution, and not every integral or root can be written down. For those, Mathilda offers numerical methods that return machine-precision answers (see the precision tutorial for what those decimals mean).

FindRoot[lhs == rhs, {x, x0}] searches for a numerical root, starting from your initial guess x0:

In[1]:= FindRoot[Cos[x] == x, {x, 1}]
Out[1]= {x -> 0.739085}

In[2]:= FindRoot[x^2 - 2 == 0, {x, 1}]
Out[2]= {x -> 1.41421}

In[1] finds the unique solution of cos x = x, a transcendental equation with no algebraic answer. In[2] converges to √2 ≈ 1.41421 — the same number Solve[x^2 - 2 == 0, x] returns exactly as Sqrt[2], here computed numerically from the guess x = 1.

FindMinimum and FindMaximum locate a local extremum, returning both the optimal value and the location that achieves it:

In[1]:= FindMinimum[x^2 - 4 x + 7, {x, 0}]
Out[1]= {3.0, {x -> 2.0}}

In[2]:= FindMaximum[-x^2 + 2 x + 1, {x, 0}]
Out[2]= {2.0, {x -> 1.0}}

In[3]:= FindMinimum[Sin[x], {x, 4}]
Out[3]= {-1.0, {x -> 4.71239}}

The parabola in In[1] has its minimum value 3 at x = 2 (its vertex), and the downward parabola in In[2] peaks at 2 when x = 1. In[3] starts near x = 4 and slides downhill to the nearest trough of sine at x ≈ 4.71239 (that is 3π/2), where the value is -1. Because the search is local, the starting point decides which extremum you find.

Where to next

You have now toured every major pillar of Mathilda's calculus: differentiation (partial, higher-order, and of abstract functions), integration by several methods, power series, limits, symbolic sums, and numerical root-finding and optimisation. Together with the algebra tutorial this covers the heart of the system's symbolic mathematics.

• For the complete reference on any function used here, browse the function documentation. The calculus and power series sections are the natural follow-ups, with the full options for D, Integrate, Series, Limit, Sum, and FindRoot.
• To revisit the guided path from the start, head back to the tutorials index.

A good habit while exploring: type ?Name at the prompt (for example ?Integrate) to read a function's built-in help string without leaving the REPL.