Differentiation Rules

How do you differentiate constants and powers of x?

The derivative of a constant is 0, and the derivative of a power of x is found by bringing the exponent down and lowering it by one: (d)/(dx)x^n=nx^n-1. The derivative functions found from the definition showed the pattern, x^2→2x and x^3→3x^2; this section proves it.

The constant rule. If f(x)=c then every difference quotient is (c-c)/(h)=0, so (d)/(dx)c=0. A horizontal line has slope 0 everywhere.

The power rule for positive whole numbers. The identity z^n-x^n=(z-x) S, where S=z^n-1+z^n-2x+…+zx^n-2+x^n-1 has n terms, can be checked by multiplying out the right side: everything cancels except z^n-x^n. By the second form of the definition, dividing by z-x leaves S, and as z→ x each of its n terms tends to x^n-1: (d)/(dx)x^n=\lim_z→ xS=nx^n-1.

The same rule for every real exponent. The formula is valid for all real n, not just positive whole numbers, wherever x^n and x^n-1 are defined; the proof for the other kinds of exponent comes with the derivative of the logarithm, and here it is used as a tool. Subtract 1 from the exponent and multiply by the old exponent:

• (d)/(dx)x^3=3x^2.
• (d)/(dx)x^2/3=\frac23x^-1/3, defined for x\ne0. At 0 the function x^2/3 has a cusp, and the formula blows up there as it should.
• (d)/(dx)x^√2=√2 x^√2-1.
• (d)/(dx)(1)/(x^4)=(d)/(dx)x^-4=-4x^-5=-(4)/(x^5).
• (d)/(dx)x^-4/3=-\frac43x^-7/3.
• (d)/(dx)x^π+1=(π+1)x^π.

The laws of exponents are what let you write roots and reciprocals as powers: √ x=x^1/2 and \frac1x^4=x^-4. That is the first move in most problems, and it turns √ x into \frac12x^-1/2=(1)/(2√ x), the formula that took a conjugate to get from the definition.

Try the tool below on any of these. It applies the rules one at a time, names each one, and checks the answer against a difference quotient.

How do the constant multiple, sum and difference rules work?

A constant factor comes out of a derivative, and the derivative of a sum or difference is the sum or difference of the derivatives: (cu)'=cu' and (u± v)'=u'± v'. The proofs are one line each, because limits of constant multiples and sums behave the same way.

Constant multiple. (cu)'=\lim_h→0(cu(x+h)-cu(x))/(h) (cu)'=c\lim_h→0(u(x+h)-u(x))/(h), which is c u'. In pictures: multiplying every height of a graph by 3 multiplies every slope by 3, so (d)/(dx)(3x^2)=3·2x=6x. With c=-1 the rule says that the derivative of -u is -u'.

Sum. The difference quotient of u+v is the sum of the difference quotients of u and of v, and the limit of a sum is the sum of the limits, so (u+v)'=u'+v'. Writing u-v as u+(-1)v gives the difference rule, and the sum rule extends to any finite number of terms.

Polynomials. Together with the power rule they differentiate any polynomial term by term. For y=x^3+\frac43x^2-5x+1, y'=3x^2+\frac43·2x-5·1+0, y'=3x^2+\frac83x-5. Every polynomial is differentiable at every x.

Where is the tangent horizontal? A horizontal tangent has slope 0, so solve y'=0. For y=x^4-2x^2+2 the derivative is y'=4x^3-4x=4x(x-1)(x+1), which is zero at x=0, 1 and -1. The horizontal tangents touch the curve at (0,2), (1,1) and (-1,1): the factoring of the derivative does the work.

The tangent line that results. With a derivative formula the tangent line at any point is two substitutions away: for y=x^3+\frac43x^2-5x+1 at x=1 the point is \left(1,-\frac53\right) and the slope is 3+\frac83-5=\frac23.

What is the derivative of e to the x?

The derivative of e^x is e^x: the natural exponential function is its own derivative, and e≈2.71828 is the one base for which that is true. Here is why. Apply the definition to f(x)=a^x with a>0: (a^x+h-a^x)/(h)=a^x·(a^h-1)/(h). Letting h→0 gives (d)/(dx)a^x=a^x· L, where L=\lim_h→0(a^h-1)/(h), and a^x was taken out because it does not depend on h. So the derivative of a^x is a constant multiple L of a^x itself, and L is a number that depends only on a. Putting x=0 shows what it is: L=f'(0), the slope of the graph of a^x where it crosses the y-axis.

That slope can be estimated from the difference quotient. With h=0.0001, L≈0.69 for a=2, about 0.92 for a=2.5 and about 1.10 for a=3. It grows with the base, so for some base between 2.5 and 3 it equals exactly 1. That base is e, the number e, and with it L=1 and (d)/(dx)e^x=e^x·1=e^x. (For a general base the value turns out to be L=\ln a; that is shown in the study of logarithms, and it is not needed here.) By the constant multiple rule, (d)/(dx)\big(c e^x\big)=c e^x, and the only functions equal to their own derivatives are these constant multiples.

Move the base in the tool below and watch L pass through 1.

A tangent through the origin. Which line through (0,0) is tangent to y=e^x? A line y=mx touching the graph at (a,e^a) has slope m=(e^a-0)/(a-0), and the slope of the curve there is e^a, so e^a=(e^a)/(a), which gives a=1 and m=e. The line is y=ex, touching the curve at (1,e).

How does the product rule work?

The derivative of a product is the first function times the derivative of the second, plus the derivative of the first times the second: (uv)'=u v'+u' v. It is not the product of the derivatives. A quick test: with u=v=x the product is x^2 with derivative 2x, while u'v'=1·1=1. The rule gives x·1+1· x=2x, as it should.

Why it is true. Let \Delta u, \Delta v and \Delta(uv) be the changes in u, v and uv when x grows by h. Adding and subtracting u(x+h)v(x) in the numerator of the difference quotient shows that \Delta(uv)=u(x+h) \Delta v+v(x) \Delta u, so (\Delta(uv))/(h)=u(x+h) (\Delta v)/(h)+v(x) (\Delta u)/(h). As h→0 the two fractions tend to v'(x) and u'(x), and u(x+h)→ u(x) because a differentiable function is continuous. So the limit is u(x)v'(x)+v(x)u'(x).

A picture. Let u and v be the sides of a rectangle of area uv. When x grows by h the sides grow by \Delta u and \Delta v, and the new area exceeds the old by two strips and a small corner: \Delta(uv)=v \Delta u+u \Delta v+\Delta u \Delta v. Divide by h: the strips give v(\Delta u)/(h) and u(\Delta v)/(h), which tend to vu' and uv', and the corner gives (\Delta u \Delta v)/(h)=\Delta u·(\Delta v)/(h)→0· v'=0, because the corner is a product of two small quantities. That is the product rule, and the tool below lets you shrink the step and watch the corner become negligible.

Examples.

• For y=(x^2+1)(x^3-3x), with u=x^2+1 and v=x^3-3x: y'=2x(x^3-3x)+(x^2+1)(3x^2-3)=5x^4-6x^2-3. As a check, expanding the brackets first gives y=x^5-2x^3-3x and the power rule gives the same y'=5x^4-6x^2-3.
• For y=x^2e^x: y'=2x e^x+x^2e^x=(x^2+2x)e^x.

When one factor is a constant the product rule reduces to the constant multiple rule, since the constant has derivative 0. For three factors, differentiate one at a time and add: (uvw)'=u'vw+uv'w+uvw'.

How does the quotient rule work?

The derivative of a quotient is the bottom times the derivative of the top, minus the top times the derivative of the bottom, all over the bottom squared: \left(\frac uv\right)'=(u'v-uv')/(v^2), wherever v\ne0. The order of the two terms in the numerator matters, because it is a difference. A way to remember it: "low d-high minus high d-low, over low-low".

Why it is true. Put (u(x+h))/(v(x+h))-(u(x))/(v(x)) over a common denominator and subtract and add u(x)v(x) in the numerator. With \Delta u and \Delta v as before and \Delta(u/v) the change in the quotient, dividing by h gives (\Delta(u/v))/(h)=(v(x)(\Delta u)/(h)-u(x)(\Delta v)/(h))/(v(x) v(x+h)). Letting h→0 turns the two fractions into u' and v' and v(x+h) into v(x), which is the rule.

Examples.

• For y=(t^2-1)/(t^3+1): y'=(2t(t^3+1)-(t^2-1)(3t^2))/((t^3+1)^2)=(-t^4+3t^2+2t)/((t^3+1)^2).
• For y=e^-x=(1)/(e^x): y'=(0· e^x-1· e^x)/((e^x)^2)=-(1)/(e^x)=-e^-x.

The reciprocal rule and negative powers. With u=1 the rule gives \left(\frac1v\right)'=-(v')/(v^2). Applied to v=x^m it recovers the power rule for negative exponents: (d)/(dx)x^-m=-(mx^m-1)/(x^2m)=-m x^-m-1.

Simplify first when you can. The quotient rule is the heaviest tool, and algebra is often lighter. For y=((x-1)(x^2-2x))/(x^4) a quotient rule would give a long expression with many terms. Instead expand the numerator and divide by x^4: y=(x^3-3x^2+2x)/(x^4)=x^-1-3x^-2+2x^-3, and then the power, multiple and sum rules give y'=-x^-2+6x^-3-6x^-4. In the same way (x^3+2x)/(x)=x^2+2 has derivative 2x with no quotient rule at all.

The tool above accepts quotients, so you can compare what it does with your own steps. It refuses things that need the chain rule, such as (x^2+1)^3, which come later.

What are second and higher-order derivatives?

The second derivative of f is the derivative of its derivative, f''=(f')', and the process can be repeated to give the third, fourth and every higher-order derivative. If f' is a function it can be differentiated again, so with a function like y=x^6: y'=6x^5, y''=30x^4, y'''=120x^3.

Notation. For y=f(x) the second derivative is written f''(x), y'', (d^2y)/(dx^2) or D^2f(x). Read (d^2y)/(dx^2) as "d squared y by dx squared", and y''' as "y triple prime". The n-th derivative is y^(n)=(d^ny)/(dx^n)=D^ny for any positive whole number n: the symbol D^2 means differentiating twice.

What it means. The first derivative is the slope of the graph, so the second derivative is the rate of change of the slope: where f''>0 the slope is increasing and the graph bends upward, and where f''<0 it bends downward. (The graph of the derivative already showed this.)

Polynomials. Each derivative lowers the degree by one, so the derivatives of y=x^3-3x^2+2 are y'=3x^2-6x, y''=6x-6, y'''=6, y^(4)=0, and every later one is 0. A polynomial has derivatives of all orders. For a polynomial of degree n with leading coefficient a_n the n-th derivative is the constant n! a_n (for x^n it is n!=n(n-1)…2·1) and the next one is 0.

The tool above shows derivatives up to the fourth for any function it accepts. Try x^5 and watch the powers fall to a constant.

How do you apply the differentiation rules in code?

A computer algebra system knows the rules: give SymPy a formula and diff applies the power, product and quotient rules for you, higher orders included. Ask it about two undefined functions u and v and it returns the rules themselves, which is a good way to see that the product and quotient rules are formulas, not tricks. It leaves the answer in the shape the rules produce, so expand, factor or simplify puts it in the form you would write: for (x^2+1)(x^3-3x) the rule gives 2x(x^3-3x)+(x^2+1)(3x^2-3), and expand turns that into 5x^4-6x^2-3, the answer found above.

import sympy as sp

x = sp.symbols('x')
u, v = sp.Function('u'), sp.Function('v')
print(sp.diff(u(x) * v(x), x))   # u(x)*Derivative(v(x), x) + v(x)*Derivative(u(x), x)
print(sp.diff(u(x) / v(x), x))   # -u(x)*Derivative(v(x), x)/v(x)**2 + Derivative(u(x), x)/v(x)

f = (x**2 + 1) * (x**3 - 3*x)
print(sp.diff(f, x))             # 2*x*(x**3 - 3*x) + (x**2 + 1)*(3*x**2 - 3)
print(sp.expand(sp.diff(f, x)))  # 5*x**4 - 6*x**2 - 3

print(sp.factor(sp.diff(x**2 * sp.exp(x), x)))               # x*(x + 2)*exp(x)
print(sp.simplify(sp.diff((x**2 - 1) / (x**3 + 1), x)))      # x*(2*x**3 - 3*x*(x**2 - 1) + 2)/(x**3 + 1)**2
print(sp.diff(x**sp.pi, x))                                  # pi*x**pi/x
print(sp.diff(x**3 - 3*x**2 + 2, x, 3), sp.diff(x**3 - 3*x**2 + 2, x, 4))   # 6 0

#include <iostream>
using namespace std;

// A polynomial is its list of coefficients: c[0] + c[1] x + c[2] x^2 + ...
int multiply(const int a[], int na, const int b[], int nb, int out[]) {
  for (int i = 0; i < na + nb - 1; i++) out[i] = 0;
  for (int i = 0; i < na; i++)
    for (int j = 0; j < nb; j++) out[i + j] += a[i] * b[j];
  return na + nb - 1;
}

void derivative(int c[], int n) {                    // c becomes c', the last coefficient 0
  for (int i = 1; i < n; i++) c[i - 1] = i * c[i];
  c[n - 1] = 0;
}

int value(const int c[], int n, int x) {             // Horner's rule
  int s = 0;
  for (int i = n - 1; i >= 0; i--) s = s * x + c[i];
  return s;
}

int main() {
  int u[3] = {1, 0, 1};                              // x^2 + 1
  int v[4] = {0, -3, 0, 1};                          // x^3 - 3x
  int f[6];
  int n = multiply(u, 3, v, 4, f);                   // f = u v
  derivative(f, n);
  for (int i = 0; i < n; i++) cout << f[i] << " ";  // -3 0 -6 0 5 0: that is 5x^4 - 6x^2 - 3
  cout << endl;

  int du[3] = {0, 2, 0};                             // u' = 2x
  int dv[4] = {-3, 0, 3, 0};                         // v' = 3x^2 - 3
  int bad = 0;
  for (int x = -5; x <= 5; x++)                      // the product rule against the derivative of the product
    if (value(f, n, x) != value(du, 3, x) * value(v, 4, x) + value(u, 3, x) * value(dv, 4, x)) bad++;
  cout << "product rule fails at " << bad << " points" << endl;

  int g[4] = {2, 0, -3, 1};                          // x^3 - 3x^2 + 2
  for (int k = 1; k <= 4; k++) {
    derivative(g, 4);
    cout << value(g, 4, 5) << " ";                   // the k-th derivative at 5
  }
  cout << endl;
  return 0;
}

/* output:
-3 0 -6 0 5 0
product rule fails at 0 points
45 24 6 0
*/

A rule is only as good as its use, so check an answer numerically. Compare the formula you found with a difference quotient at a point: if f'(x) is right, the central difference (f(x+h)-f(x-h))/(2h) with h near 10^-5 agrees with it to many digits, as in the numerical derivative of the previous article. The tool above does exactly that check for you.

Where do the differentiation rules come from?

The rules were found in the seventeenth century by the people who invented calculus, and the number e came from a question about money.

• Newton. In his work on fluxions of the 1660s and 1670s, Isaac Newton gave rules for the rate of change of sums, products, quotients and powers, including fractional powers through his binomial series.
• Leibniz. Gottfried Leibniz worked out d(xy)=x dy+y dx in manuscripts of 1675 (he first guessed that the product of differentials was the differential of the product, and then corrected himself) and stated the rules for sums, products, quotients and powers in his 1684 paper that introduced the differential calculus. His notation d^2y gave the second derivative its name.
• The number e. Jacob Bernoulli met it in 1683 while studying compound interest, as the limit of \left(1+\frac1n\right)^n. Leonhard Euler began to write it as e around 1727 and, in his Introductio in analysin infinitorum (1748), made e^x the central exponential function; the fact that it equals its own derivative is why it is called natural.
• Notation. Lagrange wrote f', f'' and f''' for successive derivatives, and Louis Arbogast introduced the operator D and its powers D^n in 1800.

What mistakes do people make with the differentiation rules?

The most common mistakes with the differentiation rules are the nine below, each with the correct statement.

Practice: test your understanding

These questions mix everything above. A wrong answer costs nothing here: read the hint and try again.

Summary: the differentiation rules at a glance

• Constant: (d)/(dx)c=0. Power: (d)/(dx)x^n=nx^n-1 for every real n where defined; write roots and reciprocals as powers first.
• Multiple, sum, difference: (cu)'=cu', (u± v)'=u'± v'; a polynomial is differentiated term by term; horizontal tangents are where y'=0.
• Exponential: (d)/(dx)a^x=a^x· L with L=f'(0); for the base e, L=1 and (d)/(dx)e^x=e^x.
• Product: (uv)'=u'v+uv', not u'v'. Quotient: \left(\frac uv\right)'=(u'v-uv')/(v^2); reciprocal \left(\frac1v\right)'=-(v')/(v^2).
• Strategy: simplify with algebra first (expand, divide); use the product or quotient rule only when needed.
• Higher order: f''=(f')', y^(n)=(d^ny)/(dx^n); f'' is the rate of change of the slope; the n-th derivative of x^n is n!.
• Code: SymPy diff applies the rules (diff(f, x, 3) for the third derivative); check with a central difference.

Glossary: key terms

Power rule

The rule that the derivative of x to the power n is n times x to the power n minus 1, valid for every real n where the powers are defined.

Constant multiple rule

The rule that the derivative of a constant times a function is the constant times the derivative of the function.

Sum rule

The rule that the derivative of a sum of differentiable functions is the sum of their derivatives; the difference rule is the same for subtraction.

Product rule

The rule that the derivative of a product uv is u times the derivative of v plus the derivative of u times v.

Quotient rule

The rule that the derivative of u over v is the bottom times the derivative of the top minus the top times the derivative of the bottom, all over the bottom squared.

Natural exponential function

The function e to the power x, whose base e is about 2.71828 and whose graph crosses the y-axis with slope 1, so that it equals its own derivative.

Higher-order derivative

A derivative of a derivative, found by differentiating repeatedly: the second derivative is the derivative of the first, the third of the second, and so on.

Reciprocal rule

The special case of the quotient rule for one over v, which says the derivative is minus the derivative of v over v squared.

Frequently asked questions

References

1. Joel Hass, Christopher Heil, Maurice D. Weir et al. (2023). Thomas' Calculus: Early Transcendentals in SI Units (15th ed.), Section 3.3: Differentiation Rules. Pearson.
2. James Stewart (2015). Calculus (8th ed.), sections 3.1 Derivatives of Polynomials and Exponential Functions, 3.2 The Product and Quotient Rules. Cengage Learning.
3. Carl B. Boyer (1949). The History of the Calculus and Its Conceptual Development. Hafner (Dover reprint, 1959).
4. Eli Maor (1994). e: The Story of a Number. Princeton University Press.
5. SymPy Development Team. SymPy tutorial: Calculus (diff). docs.sympy.org. https://docs.sympy.org/latest/tutorials/intro-tutorial/calculus.html.

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