Calculus · Eqora guide
Derivative from first principles: keep h nonzero
Build the difference quotient, cancel only for a nonzero increment and check the two-sided limit before naming a derivative.
A derivative from first principles can seem impossible when the first substitution produces 0/0. That expression is not an answer and does not mean the derivative is zero. The difference quotient compares two distinct nearby inputs; only after simplifying for a nonzero change do we study its limit. Keeping those two stages separate explains why cancellation can be valid even though substituting zero into the original quotient is not.
We will find the derivative of f(x) = x² + 3x at a = 2, identify common expansion errors, and solve a changed example independently. If you use Eqora for a homework explanation, first attempt the setup yourself, capture the complete function and requested point, inspect the steps and ask about one transition. Check the recognized notation, assumptions and result. Follow assessment rules: a learning aid cannot replace work your course requires you to complete unaided or guarantee correctness and grades.
A derivative from first principles starts with two inputs
Fix a = 2 and choose a nonzero increment h. The two inputs are 2 and 2 + h, not 2 and h. Their separation is (2 + h) − 2 = h. The corresponding output change is f(2 + h) − f(2). Dividing output change by input change gives [f(2 + h) − f(2)]/h. This is the slope of the secant joining the two graph points. A positive h approaches from the right; a negative h approaches from the left.
The University of Connecticut's definition of the derivative distinguishes this difference quotient from its limiting value. For an ordinary two-sided derivative at an interior point, nearby inputs must lie in the function's domain, and both directions must approach the same finite limit. Our polynomial is defined for every real input, so small positive and negative h are allowed. Do not assume these conditions for every function merely because the quotient has a familiar shape. State the point and domain before simplifying the algebra.
Substitute into the entire function before subtracting
Compute f(2) = 2² + 3 × 2 = 10. Next, replace every occurrence of x in f(x) by 2 + h: f(2 + h) = (2 + h)² + 3(2 + h). The square applies to the entire sum, so (2 + h)² = 4 + 4h + h². The linear term becomes 6 + 3h. Combining gives f(2 + h) = 10 + 7h + h². Keeping parentheses until both substitutions are complete protects the terms you will need next.
Now form the complete difference: f(2 + h) − f(2) = (10 + 7h + h²) − 10 = 7h + h². The subtraction applies to the whole value f(2). For a function whose value at the chosen point is negative, the outer minus would change that sign too. Writing f(2 + h) as f(2) + h would incorrectly assume that output changes by exactly the input change. That is not a general property of functions, and it would erase the behavior we are trying to measure.
Cancel a nonzero factor rather than individual terms
The quotient is (7h + h²)/h. Factor the whole numerator as h(7 + h). Because h ≠ 0, division by the common factor gives 7 + h. This equality holds for every allowed nonzero h. It does not fill in the original quotient at h = 0. The simplified expression happens to be defined there, which makes its nearby behavior easy to study. The limit uses that behavior, not a hidden permission to divide by zero.
A tempting error is to cancel h only from h² and leave the other term unchanged, or to remove symbols across an addition without factoring. Cancellation is division of a product by a common nonzero factor. Here both numerator terms contain h, which is why factoring works. If one term did not contain it, the same cancellation would be invalid. The guide to checking answers without an answer key encourages reversing an operation: multiply 7 + h by h and confirm that you recover the complete numerator 7h + h².
Take the limit and interpret the result at the point
As h approaches zero, 7 + h approaches 7 from either side. Therefore f′(2) = 7. The derivative is a slope or local rate of change at the specified input, whereas f(2) = 10 is the function's value. These numbers answer different questions. The tangent line at that point is y − 10 = 7(x − 2), or y = 7x − 4. Substituting x = 2 gives y = 10, confirming that the line passes through the correct graph point.
The original difference quotient remains undefined at h = 0. Writing “h becomes zero, so cancel zero” misstates the argument. Instead, for all nonzero h close to zero the quotient equals 7 + h, and that expression has limit 7. A general-point calculation gives [(a + h)² + 3(a + h) − (a² + 3a)]/h = 2a + 3 + h for h ≠ 0, so f′(a) = 2a + 3. Evaluating this independently derived expression at a = 2 confirms the same slope.
Use numerical changes as evidence, not as a proof
For h = 0.1, the simplified quotient is 7.1; for h = 0.01 it is 7.01. Negative changes give 6.9 and 6.99 when h = −0.1 and −0.01. The values approach 7 from opposite directions. They help interpret the limit and catch an expansion error. They do not prove that the limiting value exists for every nearby input. The algebraic expression 7 + h supplies that conclusion here because its behavior is known for all sufficiently small nonzero h.
Using extremely tiny h in a calculator is not automatically a better check. Subtracting nearly equal rounded outputs can lose useful precision before division magnifies the error. For this polynomial, evaluating the simplified quotient avoids that particular subtraction. If you only have measured values, the finite difference is an estimate influenced by the spacing and measurement error. Do not report an exact derivative from a short table unless the mathematical assumptions justify it. Distinguish what the calculation proves from what the numerical pattern merely suggests.
Capture the function, point and requested method together
For an Eqora-assisted homework session, photograph the whole question, including f(x), the point a = 2 and any instruction to use the definition. A solution obtained only by a memorized differentiation rule may be numerically correct but fail to demonstrate the requested method. Use even lighting, retain exponents and brackets, and compare the recognized task with the original. A misread square or a lost constant changes the problem before any calculus begins, so correct the transcription before studying its reasoning.
Inspect the explanation through three checkpoints: the full substitution into f(2 + h), subtraction of the complete f(2), and factoring before cancellation. Find the first checkpoint you cannot explain. The guide to learning from worked examples describes why predicting the next step is more useful than passively reading the finished chain. Cover the next line and try it yourself. The photographs show a physical ramp and careful study as analogies; they are not scaled graphs, real captured worksheets or screenshots of Eqora proving a derivative.
Ask a follow-up about the precise logical gap
Useful questions include “Why must h stay nonzero while the factor is cancelled?” and “How can an undefined quotient have a finite limit?” If the difficulty is algebraic, ask “Which term appears when (2 + h)² is expanded, and why?” These questions identify a transition that can be justified and checked. Asking for another final answer may leave the same confusion intact, even when the new explanation uses different words or a more polished arrangement of equations.
After reading a response, close it and explain the step on paper. Mark the nonzero condition beside the cancellation and the limit symbol beside the final passage. If Eqora supplies a result, verify it by expansion and recombination rather than accepting confident language as evidence. It is a learning aid, not a guarantee of accuracy or grades. Confirm the original task, notation, domain and conclusion. If your course uses a different convention for endpoint derivatives, ask your teacher to clarify the convention before applying it to a boundary example.
Check a case where the two directions disagree
Consider p(x) = |x| at zero. Its difference quotient is (|h| − 0)/h. For positive h this equals 1, and for negative h it equals −1. The two one-sided limits differ, so there is no ordinary two-sided derivative at zero. The function is defined and continuous there, but that does not force differentiability. A symmetric picture or a calculator output at the center cannot replace examining both directions. The corner is a genuine mathematical difference from our polynomial example.
In particular, [p(h) − p(−h)]/(2h) equals zero for every nonzero h. That symmetric difference still does not make p′(0) equal to zero: it is a different quotient, and its limit is insufficient to establish the ordinary derivative. This example is a useful warning about choosing a numerical check without understanding what it measures. A proposed method must match the derivative definition being used. Always preserve the base point and both permitted directions rather than replacing the original problem with a more convenient symmetric calculation.
Solve an altered function independently
Try g(x) = 3x² − x at a = −1 with the explanation covered. First g(−1) = 4. Expanding the shifted value gives g(−1 + h) = 3(−1 + h)² − (−1 + h) = 4 − 7h + 3h². Subtracting 4 and dividing by nonzero h yields −7 + 3h. Its two-sided limit is −7, so g′(−1) = −7. Reverse the simplification by multiplying by h, then check that the expanded numerator returns.
The general expression g′(a) = 6a − 1 confirms −7 at a = −1. Explain the sign: a negative derivative indicates decreasing output for small positive input changes near that point; it is not a negative function value or distance. If you can set up the quotient, justify the nonzero cancellation and separate the limit from substitution, you have a method you can reuse. Finish with the requested point, the exact derivative and a check you performed independently, rather than only a copied sequence of symbols.
