Limits and Derivatives · Lesson 6 of 8
Derivative Rules and Algebraic Functions
“Derive the power rule and combine differentiation rules to handle algebraic expressions efficiently.”
• Apply sum, difference, constant-multiple, product, and quotient rules correctly. • Derive the power rule using binomial expansion and by induction. • Differentiate polynomials and evaluate the resulting derivative at a point. • Choose between expanding, simplifying, and applying a rule. • Keep constants, variable inputs, and domain restrictions distinct.
Rules built on the definition
First principles explain what a derivative means, but repeating the full difference quotient for every expression would be lengthy. Once the component derivatives exist, rules let us combine them. These rules describe derivatives of sums, differences, products, and quotients.
The sum and difference rules closely resemble the corresponding limit rules. The product and quotient rules are different from simply multiplying or dividing the derivatives. A changing product depends on changes in both its factors, so its rate must include both contributions.
Here u and v are functions of x, and their primes mean derivatives with respect to x. A fixed constant has derivative zero. Thus the constant-multiple rule is also a special case of the product rule: differentiating c·u gives 0·u+c·u′.
The two terms say: hold v at its current value while differentiating u, then hold u at its current value while differentiating v, and add the results. This is also called the Leibniz rule. The full factor is retained in each term; it is not replaced by its derivative everywhere.
The derivative of a product is usually not u′v′, and the derivative of a quotient is usually not u′/v′. For example x·x has derivative 2x, not 1; x/x has derivative 0 where defined, not 1/1.
Simple checks make the rules concrete
We already know that the derivative of x is 1 and the derivative of a constant is zero. These two facts are enough to check the rules against first-principles results. Agreement between two valid methods is a useful way to catch algebraic errors.
Problem
Use the rules to differentiate 10x and x².
- 1.For 10x, treat 10 as a constant factor: the derivative is 10×1=10.
- 2.Alternatively, write 10x as a sum of ten copies of x; the sum rule gives ten copies of 1.
- 3.For x²=x·x, the product rule gives 1·x+x·1=2x.
- 4.Both answers agree with the earlier first-principles calculations.
There is also a direct reason for the product rule. In the output change of u(x)v(x), add and subtract u(x)v(x+h). After grouping, divide by h. One term contains the difference quotient of u multiplied by v(x+h); the other contains u(x) multiplied by the difference quotient of v.
As h→0 the derivative quotients tend to u′(x) and v′(x). A differentiable function approaches its point value: for example, v(x+h)−v(x) is h times a quotient with a finite limit, so that difference tends to zero. The two limiting contributions are therefore u′v and uv′. The rules can be used without reproducing this argument each time.
Proving the power rule
Let f(x)=xⁿ, where n is a positive integer. In the first-principles definition, expand (x+h)ⁿ using the binomial theorem. The first term xⁿ cancels, the next term is nxⁿ⁻¹h, and every later term contains at least h². After dividing by h, only the term nxⁿ⁻¹ survives the limit.
An alternative proof uses induction and the product rule. The case n=1 is already known. Suppose the result holds for exponent n−1. Write xⁿ=x·xⁿ⁻¹. Its derivative is 1·xⁿ⁻¹+x·(n−1)xⁿ⁻², which equals nxⁿ⁻¹. This proves the next integer case.
The chapter states that the rule extends to real powers wherever the real-valued function and derivative are defined, without proving the general extension. For a negative power, x=0 is excluded. For fractional powers, check the real domain; working at x>0 avoids the usual endpoint and real-root difficulties.
Problem
Differentiate 1/x² for x≠0.
- 1.Write u=1 and v=x² in the quotient rule. Their derivatives are 0 and 2x.
- 2.The result is (0· x²-1·2x)/x⁴=-2/x³.
- 3.Writing the function as x⁻² gives the same result by the extended power rule. Both methods preserve x≠0.
Polynomials are differentiated term by term
A polynomial is a sum of constant multiples of powers of x. The sum, constant-multiple, and power rules therefore combine directly. Multiply each non-constant coefficient by its exponent and reduce that exponent by one. The constant term disappears because its derivative is zero.
Problem
Differentiate 6x¹⁰⁰-x⁵⁵+x.
- 1.The first term gives 6×100x⁹⁹=600x⁹⁹.
- 2.The second gives −55x⁵⁴, and the derivative of x is 1.
- 3.Adding the three derivatives gives 600x⁹⁹-55x⁵⁴+1. The size of an exponent does not change the rule.
Problem
For f(x)=1+x+x²+⋯+x⁵⁰, find f′(1).
- 1.Differentiate term by term to obtain f'(x)=1+2x+3x²+⋯+50x⁴⁹.
- 2.Now substitute x=1 into the derivative, giving 1+2+⋯+50.
- 3.Pairing the first and last terms gives 25 pairs each with sum 51. Thus f′(1)=25×51=1275.
- 4.Substituting x=1 into f first would produce a number and lose the varying function that must be differentiated.
Rational functions: simplify or use the quotient rule
The quotient rule provides a general method, but simplifying a function first may reduce the work. The answer should agree wherever the original function is defined. Always retain restrictions from the original denominator, even if an algebraic cancellation removes that denominator later.
Problem
Differentiate (x+1)/x for x≠0.
- 1.With u=x+1 and v=x, u′=1 and v′=1.
- 2.The quotient rule gives [1· x-(x+1)·1]/x²=-1/x².
- 3.Alternatively, simplify the function to 1+1/x. The derivative of the constant is zero and the reciprocal derivative is −1/x².
- 4.The domain remains x≠0 in either calculation.
Problem
Differentiate f(x)=(2x+3)/(x-2) from first principles.
- 1.Require x≠2 and, during the nearby calculation, x+h≠2.
- 2.After combining the two function values over a common denominator, the difference-quotient numerator is (2x+2h+3)(x-2)-(2x+3)(x+h-2).
- 3.Separate the terms involving h. The x-dependent base products cancel, leaving h[2(x−2)−(2x+3)]=−7h.
- 4.Cancel the non-zero h. The quotient becomes -7/[(x-2)(x+h-2)].
- 5.Its limit is -7/(x-2)². The quotient rule also gives [2(x-2)-(2x+3)]/(x-2)², confirming the result.
Problem
Find the derivative of f(x)=x+1/x from first principles.
- 1.The function requires x≠0. Its difference quotient is [h+1/(x+h)-1/x]/h.
- 2.Combine the reciprocal difference as -h/[x(x+h)]. The complete numerator is h[1-1/(x(x+h))].
- 3.Cancel h and take the limit to get 1-1/x².
- 4.The sum rule gives the same answer from the two component derivatives, and zero remains excluded.
A fixed number inside a power
Some exercises use a power of ax+b, with a and b fixed. The factor a affects how fast the base changes. For positive integer n, this can be derived with the same binomial reasoning, so we do not need to introduce an additional general differentiation rule.
Replace x by x+h: the base ax+b increases by ah. In the binomial expansion of [(ax+b)+ah]ⁿ, the first surviving difference term is n(ax+b)ⁿ⁻¹ah. Dividing by h and taking the limit gives na(ax+b)ⁿ⁻¹. Negative integer powers can be treated as reciprocals using the quotient rule; the constant case has derivative zero.
Problem
Differentiate (ax²+b)² for fixed constants a,b.
- 1.Expand first: (ax²+b)²=a²x⁴+2abx²+b².
- 2.Treat a and b as constants. Differentiation gives 4a²x³+4abx.
- 3.Factor the answer as 4ax(ax²+b). Alternatively, write the original square as a product of two identical factors and use the product rule.
Quiz
Which is the product rule?
What is the derivative of 6x¹⁰⁰?
Which numerator belongs in the quotient rule for u/v?
What is the derivative of x³−4x+9?
For f(x)=(x+1)/x, which input remains excluded after simplification?
What is the derivative of (2x+3)³?
Practice Problems
- For f(x)=x¹⁰⁰/100+x⁹⁹/99+⋯+x²/2+x+1, prove that f′(1)=100f′(0).
- Differentiate xⁿ+ax⁽ⁿ⁻¹⁾+a²x⁽ⁿ⁻²⁾+⋯+a⁽ⁿ⁻¹⁾x+aⁿ for a fixed a and positive integer n.
- Differentiate each of (x-a)(x-b), (ax²+b)², and (x-a)/(x-b), treating a,b as constants.
- Differentiate (xⁿ-aⁿ)/(x-a) for positive integer n. State the original restriction x≠a, even if you use factorisation first.
- Differentiate each of 2x-3/4, (5x³+3x-1)(x-1), x⁽⁻³⁾(5+3x), x⁵(3-6x⁽⁻⁹⁾), x⁽⁻⁴⁾(3-4x⁽⁻⁵⁾), and 2/(x+1)-x²/(3x-1). State restrictions.
- Differentiate x+a for a fixed constant a. Explain why a has derivative zero.
- Differentiate (px+q)(r/x+s) for fixed non-zero constants and x≠0.
- Differentiate (ax+b)(cx+d)² by product rule or expansion.
- Differentiate (ax+b)/(cx+d). Show that the numerator of the result simplifies to ad−bc.
- Differentiate (1+1/x)/(1-1/x), retaining x≠0 and x≠1 after simplifying.
- Differentiate 1/(ax²+bx+c) where the denominator is non-zero.
- Differentiate (ax+b)/(px²+qx+r) where the denominator is non-zero.
- Differentiate (px²+qx+r)/(ax+b) where ax+b is non-zero.
- Differentiate 4√(x)-2 for x>0. Explain why the derivative formula cannot be evaluated at x=0.
- Differentiate (ax+b)ⁿ for integer n, using binomial reasoning for positive n and reciprocal reasoning for negative n.
- Differentiate (ax+b)ⁿ(cx+d)ᵐ for integers m,n by combining the preceding result with the product rule. State any restrictions from negative powers.
Key Takeaways
• Sums, differences, and constant multiples differentiate term by term. • A product needs two contributions; a quotient uses u′v−uv′ over v². • The integer power rule follows from the binomial theorem or induction. • Polynomial differentiation combines the power and sum rules. • Equivalent methods can simplify work, but cancelled factors do not erase original domain restrictions.