Differentiation

01 / UNDERSTAND & APPLY

1. The derivative is a limit, not a small finite quotient

At an interior point a of the domain, f′(a)=lim_{h→0}[f(a+h)−f(a)]/h when this two-sided limit exists as a finite real number. The increments h are nonzero during the quotient calculation. Setting h=0 before simplifying creates 0/0 rather than a derivative.

The quotient is an average rate of change over an interval. Its limit is the instantaneous rate at a. If f has units of distance and the input has units of time, f′ has units of distance per time. A finite derivative also gives the tangent-line slope when the graph is represented as y=f(x).

Differentiability means f(a+h)=f(a)+f′(a)h+r(h), where r(h)/h→0. Thus the derivative is the coefficient of the best first-order local change. This statement explains why a tangent-line approximation improves as the input moves closer to the point, while also showing that the approximation is not usually exact.

A derivative at one point is a number; the derivative function f′ assigns such numbers at the points where they exist. Its domain may be smaller than the original domain. For f(x)=√x, the formula f′(x)=1/(2√x) holds for x>0. The endpoint x=0 has no finite right derivative, even though f itself is defined there.

02 / UNDERSTAND & APPLY

2. Linearity and powers provide the basic pieces

The derivative of a constant is zero. For differentiable f,g and constants α,β, the derivative of αf+βg is αf′+βg′. This follows by distributing the difference quotient and using the laws for finite limits. Constants multiplying functions stay outside the derivative; additive constants disappear.

For a positive integer n, the derivative of xⁿ is nxⁿ⁻¹. Expand (x+h)ⁿ by the binomial theorem, subtract xⁿ and divide by h. The first surviving term is nxⁿ⁻¹, while all other terms contain positive powers of h and vanish in the limit.

The same formula extends to real powers on x>0. For integer or suitable rational powers it may also hold on negative inputs where the real-valued function is defined and differentiable. Always state the domain rather than applying a real-power rule across a branch or an excluded zero.

Simplifying a formula before differentiating can reduce work, but it cannot fill a hole in the original domain. The function (x²−1)/(x−1) equals x+1 only for x≠1; its derivative is 1 on that domain. A separately defined continuous extension has a derivative at 1, but the original function does not.

03 / UNDERSTAND & APPLY

3. Products and quotients have interaction terms

For differentiable u,v, the product rule is (uv)′=u′v+uv′. In a small change, both factors vary. The first-order effects come from changing one factor at a time; the product of the two small changes is of higher order and does not survive after division by the input increment and passage to the limit.

A limit derivation inserts and subtracts u(x+h)v(x) in the product difference. After dividing by h, one term contains the difference quotient of u times v(x), and the other contains u(x+h) times the difference quotient of v. Differentiability implies continuity, so u(x+h)→u(x), giving the stated rule.

For v≠0, write u/v=u·v⁻¹ or differentiate v·(u/v)=u. Solving for the derivative gives (u/v)′=(u′v−uv′)/v². The subtraction order is numerator derivative times denominator minus numerator times denominator derivative. Reversing it changes the sign.

The derivative of a product is generally not the product of the derivatives, and the derivative of a quotient is generally not their quotient. For instance, (x·x)′=2x, whereas x′·x′=1. Choose the rule based on the outermost operation in the original expression.

04 / UNDERSTAND & APPLY

4. Composition multiplies local rates

If g is differentiable at x and f is differentiable at g(x), the chain rule gives (f∘g)′(x)=f′(g(x))g′(x). The outer function is differentiated at the inner value, then multiplied by the derivative of the inner function. Simply differentiating the outer shape omits a rate conversion.

For a three-stage composition, repeat the process through every layer. For y=sin((x²+1)³), the layers are sine, cube and x²+1. The result is cos((x²+1)³)·3(x²+1)²·2x. Writing temporary variables can make the nesting visible and prevent missed factors.

The notation dy/dx=(dy/du)(du/dx) is a useful mnemonic when y=f(u),u=g(x), but the chain rule is a theorem about limits, not unrestricted fraction cancellation. Its validity does not require g′(x) to be nonzero. By contrast, solving for dx/dy through a reciprocal does require a nonzero derivative and an appropriate local inverse.

When several operations appear, start with the outermost one and recurse. In a product containing a composite function, use the product rule first, then the chain rule inside the affected term. A clean factored answer is often easier to inspect than a large expansion.

05 / UNDERSTAND & APPLY

5. Standard derivatives carry hidden conventions

With angles measured in radians, (sin x)′=cos x and (cos x)′=−sin x. The quotient rule then gives (tan x)′=sec²x where cos x≠0 and (cot x)′=−csc²x where sin x≠0. Similarly (sec x)′=sec x tan x and (csc x)′=−csc x cot x on their domains.

Radians are essential to these simple coefficients. If a numerical angle x is measured in degrees, sin of that angle is sin(πx/180) in radian notation, whose derivative with respect to x is (π/180)cos(πx/180). A missing unit conversion is a chain-rule error.

For a>0, (aˣ)′=aˣ ln a. The special base e gives (eˣ)′=eˣ; base 1 gives the constant function with derivative zero. For x>0, (ln x)′=1/x and (logₐx)′=1/(x ln a) when a>0,a≠1.

For a differentiable nonzero u(x), d/dx ln|u(x)|=u′(x)/u(x) on intervals where u remains nonzero. The absolute value permits negative u without attempting a real logarithm of a negative number. It does not make u=0 admissible.

06 / UNDERSTAND & APPLY

6. Differentiate an inverse through its defining identity

If y=f⁻¹(x) on an appropriate local inverse branch, then f(y)=x. Differentiating gives f′(y)y′=1, hence (f⁻¹)′(x)=1/f′(f⁻¹(x)), provided the denominator is nonzero and the inverse is differentiable under the usual local hypotheses. A continuous derivative nonzero near the point is a standard sufficient setting.

For y=arcsin x in [−π/2,π/2], sin y=x. On −1<x<1, cos y=√(1−x²)>0, giving y′=1/√(1−x²). For arccos x in [0,π], differentiation gives −1/√(1−x²) on the same open interval.

For y=arctan x in (−π/2,π/2), tan y=x and sec²y=1+x², so y′=1/(1+x²) for all real x. Under the common arccot branch (0,π), its derivative is −1/(1+x²). State the branch convention when using inverse trigonometric notation.

An inverse can exist at a point while failing to have a finite derivative there. The inverse of x³ is the real cube root, whose slope is unbounded at zero because the original derivative is zero. Likewise arcsin is defined at ±1 but has no finite ordinary derivative there.

07 / UNDERSTAND & APPLY

7. Dependent variables also change

An equation F(x,y)=0 may describe y locally as a function of x even without an explicit solved formula. Differentiate every occurrence of y using the chain rule. Thus d(y²)/dx=2yy′ and d(xy)/dx=y+xy′. Solving the resulting equation for y′ is valid where its coefficient is nonzero.

For a continuously differentiable F, the condition F_y≠0 at a point on the curve guarantees a local differentiable y(x), with y′=−F_x/F_y. If F_y=0, that formula cannot be used to declare a finite slope; the point may have a vertical tangent, another local description or a singularity requiring further analysis.

For a parametrized curve x=x(t),y=y(t) with x′(t)≠0 and appropriate smoothness, dy/dx=y′(t)/x′(t). A second derivative is d²y/dx²=[d/dt(dy/dx)]/x′(t). It is not generally y″(t)/x″(t). If x′ vanishes, return to the curve or examine a limiting ratio rather than divide by zero.

Implicit and parametric differentiation describe local branches. A curve can revisit a point with different parameter values and different tangents. Check the specified point and parameter before substituting into a derivative formula.

08 / UNDERSTAND & APPLY

8. Variable powers are not ordinary power-rule problems

For y=u(x)^{v(x)} with u(x)>0, take natural logarithms: ln y=v ln u. Differentiate to obtain y′/y=v′ln u+v u′/u. Multiplying back by y gives the full derivative. Both the base and exponent may contribute.

The ordinary power rule rxʳ⁻¹ treats r as constant, while the exponential rule aˣ ln a treats a as constant. Neither alone applies to xˣ. The positive-base restriction gives a single smooth real-valued meaning to arbitrary variable real exponents.

Logarithmic differentiation also simplifies products and quotients with many factors. On an interval where a differentiable y is nonzero, ln|y| has derivative y′/y. Use absolute values for signed factors where appropriate, and inspect zeros separately because dividing by y can hide valid derivative values there.

09 / UNDERSTAND & APPLY

9. Differentiate the result, not the notation

The second derivative f″ is the derivative of f′ wherever it exists. It measures how the first-order rate changes. In motion, position has velocity as its first time derivative and acceleration as its second; their units differ by another factor of inverse time.

Higher derivatives do not require the original function to be a polynomial, and differentiability once does not guarantee differentiability twice. For example, f(x)=x|x| has f′(0)=0, but its derivative is 2|x|, which has no derivative at zero.

For a product of sufficiently differentiable functions, repeated product differentiation gives (uv)″=u″v+2u′v′+uv″. More generally the nth derivative sums C(n,k)u^(k)v^(n−k) over k=0,…,n. The binomial coefficients count how many sequences of differentiations place k derivatives on the first factor.

Check a derivative by a second algebraic route, dimensional consistency or a numerical difference quotient at a safe interior point. A numerical check can catch an omitted factor but is not a proof of a formula on the whole domain. Avoid points where roundoff or a singularity overwhelms the small difference being divided.

10 / UNDERSTAND & APPLY

10. Original practice with complete reasoning

All trigonometric arguments are in radians. Give derivative domains and explain any point where a formula cannot simply be substituted.

11 / UNDERSTAND & APPLY

11. A structured derivative is easier to trust

Begin with the function domain and the outermost operation. Decide whether the expression is a sum, product, quotient or composition, then apply the corresponding rule layer by layer. Keep factored expressions when they expose missing factors more clearly than expansion would.

In implicit work, every y depends on x. In parametric work, the conversion from t-rate to x-rate is needed each time, including for the second derivative. In logarithmic work, positivity or nonzero conditions justify the logarithm and division steps.

The major warning signs are a product differentiated as a product of derivatives, a missing inner derivative, a sign reversal in the quotient rule, degrees treated as radians, a negative standard inverse-square-root branch, or a canceled factor whose excluded input has silently returned.

Finally distinguish a formula from existence at a particular point. A derivative can fail because of a corner, an unbounded slope, oscillation or an undefined original value. The difference quotient and the local domain remain the final tests when a memorized rule does not apply.