Power, chain, product and quotient rules
Choose differentiation rules from the expression structure, retain real domains, and verify combined derivatives by another form.
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Begin with the function’s real domain
Differentiation describes local change at an allowed input. Before choosing a rule, record where the original real function exists and where a finite derivative can exist. A readable derivative formula cannot restore an excluded input.
For y = xⁿ, the power rule gives dy/dx = nxⁿ⁻¹ wherever the real function is differentiable. Constants differentiate to zero; constant multiples, sums and differences differentiate term by term.
For example, on x > 0,
so
The reciprocal excludes zero, and the square root requires a nonnegative input. Together they give the original domain x > 0. Rewriting with powers helps the calculation but does not remove those conditions.
Different functions can fail to have a finite derivative for different reasons. The function 1/x has no point at zero. The function square root of x has a point at zero, but no finite tangent gradient there. The real cube-root function cube root of x exists for every real x, with derivative 1/(3x^(2/3)) for x ≠ 0; that derivative is not finite at zero either. For negative x in this last expression, x^(2/3) = (cube root of x)².
A chain combines an outer and an inner change
For y = f(u) with u = g(x), the chain rule is
where the required functions and derivatives exist. The outer derivative is evaluated at the inner output; the inner derivative tells how quickly that input changes with x.
For y = (3x² + 1)⁵, put u = 3x² + 1. Then dy/du = 5u⁴ and du/dx = 6x, giving
Omitting 6x treats the inner input as if it changed at unit rate. At x = 0, the inner rate is zero and the derivative is zero, even though the original function value is one.
Check your understanding 1 Differentiate the inside too
For y = square root of (2x + 5), state the maximal real domain, find the derivative where it is finite, and evaluate it at x = 2. What happens at the boundary input?
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The function requires 2x + 5 ≥ 0, so its domain is [-5/2,∞). For x > -5/2, the chain rule gives y' = (1/2)(2x + 5)^(-1/2) · 2 = 1/square root of (2x + 5). At x = 2, this is 1/3.
At x = -5/2, the function value is zero, but the derivative is not finite. The derivative’s domain is therefore smaller than the function’s domain. The inner derivative cancels the numerical half; it still has to be included.
The conditions of the rule matter when the inner output reaches a boundary of the outer function. A zero inner derivative cannot simply be multiplied by an undefined outer derivative.
Check your understanding 2 A square root does not undo every square
For y = square root of x² on the real line, identify the function and give its derivative on each side of zero. Does it have a derivative at zero? Explain why replacing it by x gives a wrong answer for negative inputs.
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The function is |x|: it equals x for x ≥ 0 and -x for x < 0. Its derivative is 1 for x > 0 and -1 for x < 0. At zero, the right and left gradients disagree, so the function is not differentiable there.
For nonzero x, the chain calculation gives x/square root of x² = x/|x|, agreeing with those two pieces. At zero it gives no derivative. The function itself still exists there; this is a corner, not a removed input.
A product has two changing contributions
For y = u(x)v(x),
Each factor changes while the other supplies its current value. Differentiating both and multiplying u'v' misses those two contributions.
For y = (x² + 1)(x³-2),
An alternative is to expand first: y = x⁵ + x³-2x²-2, so y' = 5x⁴ + 3x²-4x. Expanding the product-rule result gives the same derivative. At x = 1, both correct forms give 4, whereas the incorrect product u'v' = 6x³ would give 6.
Not every expression needs the product rule. A constant multiple such as 7x³ differentiates to 21x² directly; the constant’s derivative contributes zero if you choose to write the product rule out.
A quotient keeps its denominator condition
Write u = yv and apply the product rule: u' = y'v + yv'. Substituting y = u/v and rearranging gives
The subtraction applies to the whole uv' product. The result is not u'/v', and the original denominator condition remains part of the answer.
For y = (x² + 1)/(x-1),
Division provides another method: y = x + 1 + 2/(x-1) on the same domain, so y' = 1-2/(x-1)². Combining that expression over a common denominator recovers the quotient-rule result. Either form is valid; simplifying never fills an original hole or pole.
Check your understanding 3 Subtract the complete product
Differentiate y = (x² + 3)/(2x-1) and state its domain. A proposed numerator after applying the quotient rule is 2x(2x-1)-2x² + 6. Identify the error and check your derivative at x = 1.
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The domain excludes x = 1/2. The correct numerator is 2x(2x-1)-2(x² + 3), so
The proposed numerator failed to subtract the constant part of the complete product: -2(x² + 3) = -2x²-6, not -2x² + 6. At x = 1, the correct derivative is -6.
Choose the structure before simplifying
Several rules may be needed in one expression. For y = x² square root of (2x + 1), the outer structure is a product, and one factor is itself a composition. On x > -1/2,
The original function also exists at x = -1/2, but does not have a finite derivative there. Applying only the power rule to the visible square root would miss the inner factor; differentiating only one factor would miss part of the product’s change.
Try it yourself 1
Check all the zero gradients
Problem
A function is m(x) = (x²-1)²/(x² + 1). Find its real domain and differentiate it, showing where each rule is used. Verify your derivative by first rewriting the original function as x²-3 + 4/(x² + 1). Find every real input with zero gradient. A claim lists only x = ±1 because those make the original numerator zero: decide whether it is complete.
Write your answer, then compare it with the solution
Hints
Hint 1
Keep the function and its derivative separate
Track the outer quotient and the squared inner expression. When solving for a zero gradient, check every factor of the derivative rather than dividing away a possible zero.
Show full solution
Differentiate the grouped expression
Method
Use the chain rule for the numerator within the quotient rule.Reason
The squared numerator and the denominator both change with x.Working
Answer
The denominator x² + 1 is positive for every real x, so the domain is R. The numerator derivative is 4x(x²-1). Thus m' = [4x(x²-1)(x² + 1)-2x(x²-1)²]/(x² + 1)² = 2x(x²-1)(x² + 3)/(x² + 1)².Verify through a different form
Method
Differentiate x²-3 + 4(x² + 1)⁻¹.Reason
Equivalent forms on the original domain must give the same derivative.Working
Answer
This gives m' = 2x-8x/(x² + 1)². Combining the terms yields 2x[(x² + 1)²-4]/(x² + 1)², which factors into the first result.Retain every allowed zero-gradient input
Method
Set the factored numerator of the derivative equal to zero.Reason
The derivative, not the original function value, determines the tangent gradient.Working
Answer
The factor x² + 3 cannot be zero for real x, and the denominator never vanishes. The complete set is x = -1,0,1. The claim omitted 0, where m(0) = 1 but m'(0) = 0. Dividing the derivative equation by x would lose this valid input.
Quick check
A correct derivative may remain grouped, expanded or factored; check equivalence on the original domain. The three zero-gradient inputs are all allowed. They are candidates for later stationary-point analysis, not classifications obtained merely by setting a numerator to zero.
Record the original domain, identify the expression’s outer structure, then follow every changing inner part. Keep the full grouping of products and quotients. Use an independent algebraic form or a direct value check where it can expose a missing factor or sign.
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