Unit notes

A composition f(g(x)) applies g first and then f. An implicit equation relates x and y without necessarily solving for y. When y depends on x, differentiating a y-expression requires a factor of dy/dx.

For a composition f(g(x)), changing x changes g, which then changes f. The derivative multiplies the outer derivative evaluated at the inner expression by the inner derivative. Work from the outside inward and preserve the inner expression until its own derivative is taken.

A common error is differentiating only the outer function. Another is evaluating the outer derivative at x rather than at g(x). Tables can supply the needed values: to find (f∘g)′(a), locate g(a), then f′ at that output, and multiply by g′(a).

An equation can define y through a relationship with x without an explicit y = f(x) formula. Differentiate both sides with respect to x, attaching y′ whenever the chain rule differentiates a y-dependent expression. Then solve for y′ if possible.

Substitute a point only after differentiating unless the requested reasoning explicitly permits another method. Substituting constant coordinate values too early erases the variable relationship. A zero denominator in a slope expression may indicate a vertical tangent, but examine numerator and local curve behavior before concluding.

If f is one-to-one near a point and f′ is nonzero there, the inverse’s derivative is the reciprocal of the original derivative at the corresponding input. The evaluation point matters: an inverse input is an original output.

Inverse trigonometric derivatives follow from this relationship with appropriate restricted domains. For example, d/dx(arctan x) = 1(1+x²), while d/dx(arcsin x) = 1√(1−x²) for −1 < x < 1. A composite argument requires another chain-rule factor.

For u(x) > 0, d/dx ln u = u′/u. More generally, ln|u| has the same derivative wherever u is nonzero and differentiable. For a positive constant base b ≠ 1, the derivative of b^x is b^x ln b, and the derivative of log_b x is 1(x ln b).

Distinguish a variable base raised to a constant power from a constant base raised to a variable exponent. Expressions such as x^x combine both dependencies and can be handled by logarithmic differentiation on a suitable positive domain. State that domain before taking logs.

The second derivative is the derivative of the first derivative. In motion, it is acceleration when the original function is position. A third derivative would describe the rate of change of acceleration, but interpret only what the problem asks.

When an implicit first derivative contains x and y, differentiating again requires the chain rule and often the product or quotient rule. Replace y′ using the first-derivative relationship only after correctly differentiating. Do not treat y as a constant in the second step.

For y=(3x²+1)⁴, the outer operation is raising something to the fourth power and the inner expression is 3x²+1. Differentiate the outer operation while leaving the inner expression intact, then multiply by the derivative of the inner expression: y′=4(3x²+1)³(6x). The inner expression is not replaced by its derivative inside the power.

Nested compositions may require several factors. In sin((x²+1)³), differentiate sine, then the cube, then x²+1. Name intermediate variables if the nesting is hard to see. A product can also contain compositions, so product and chain rules can be needed in the same line. Check each factor against a particular layer rather than memorizing a visual pattern.

Differentiate both sides with respect to x, treating y as a function of x. The derivative of y² is 2y y′, and the derivative of xy is y+xy′ by the product rule. Collect all terms containing y′ on one side, factor y′, and divide only after identifying where the denominator is nonzero.

A requested tangent slope at a point requires both coordinates to satisfy the original equation. Substitute them into the derivative expression after differentiating. A zero denominator can signal a vertical tangent or a more complicated situation; inspect the original relation and numerator rather than declaring every such point a vertical tangent. For a second derivative, differentiate the first-derivative relation again and retain the dependence of y on x.

If f is invertible near a and f′(a)≠0, then the inverse derivative at f(a) is 1f′(a). The input of the inverse is an output of the original function. In a table, first find the row where f(a) equals the desired inverse input, then use the slope from that row. Confusing the row input with its output is a common error.

Inverse trigonometric functions use restricted ranges to make them single-valued. The derivative of arcsin x involves √(1−x²), while the derivative of arctan x involves 1+x². A composition such as arctan(2x) also requires the inner derivative. Check the function’s domain and where the derivative formula is defined; a formula does not remove endpoint restrictions.

For f(g(x)), differentiate the outside function at the inside value, then multiply by g′(x). Naming the inner expression u can make the dependency visible. In sin(x²), the cosine is evaluated at x² and multiplied by 2x; differentiating only the outer function misses how fast its input changes.

Nested expressions may require several chain-rule factors. Work from the outside inward, preserving each unchanged inner expression until its turn. A product inside a composition still needs its own product rule. Parentheses are part of the reasoning, not decoration.

A units check can confirm the chain structure: if y changes with u and u changes with x, then (dy/du)(du/dx) has units of y per x. This same relationship later supports related rates and substitution in integration.

An equation such as x²+y²=25 describes a relation. Differentiate both sides with respect to x, writing dy/dx whenever a y-dependent term is differentiated. The derivative of y² is 2y·dy/dx; treating y as a constant changes the problem.

Collect derivative terms and solve for dy/dx. Evaluate at a point only after obtaining the derivative relation unless a carefully organized equivalent method is used. Verify that the point lies on the original curve. A zero denominator can signal a vertical tangent or another feature requiring further analysis, not automatically a horizontal tangent.

For a second derivative, differentiate the first derivative while continuing to treat y as a function of x. Substitute the first-derivative expression when needed. The presence of dy/dx in the intermediate calculation is expected.

If g=f^-1, then g′(a)=1f′(g(a)) when the relevant derivative exists and is nonzero. First find the original input that produces a. The slope is reciprocated at corresponding points, not at the same numerical x-coordinate by default.

Inverse trigonometric derivatives arise from restricted branches. Their domains matter: arcsine requires inputs between −1 and 1 and has derivative behavior that becomes unbounded at the endpoints. A chain-rule factor is still required when its argument is a function of x.

Logarithmic differentiation can simplify products, quotients, or variable powers when expressions are positive or absolute values are handled appropriately. Taking logarithms does not eliminate the chain rule: differentiating ln y gives y′/y. Finish by solving for y′ and replacing y with the original expression.

For h(x)=f(g(x)), begin by finding g at the requested input. Then look up f′ at that output, and multiply by g′ at the original input. A common mistake uses f′ at the original input as well, ignoring the composition. A small dependency diagram can prevent that substitution error.

For h(x)=f(x)g(x), the two terms use function values and derivatives at the same input. A table must provide f, g, f′, and g′ there. For a quotient, check that the denominator’s function value is nonzero. The existence of numbers in a table does not establish an unstated derivative elsewhere.

When a problem gives an inverse, first locate the row whose f-value equals the inverse input. This is a search for a corresponding point. Only then reciprocate the derivative. Keeping a written chain of input → inner output → outer derivative makes table problems as systematic as symbolic ones.

A second derivative differentiates the entire first derivative. If a chain-rule result is a product, the next pass needs the product rule as well as any chain rule. For y=sin(x²), y′=2x cos(x²), and y″ includes both 2cos(x²) and −4x²sin(x²). Omitting the derivative of 2x loses one term.

With implicit relations, every occurrence of y remains dependent on x during the second pass. Substitute y′ only after preserving all dependency factors. The result may be simplified using the original relation, which can reveal a compact expression and improve sign analysis.

An inverse derivative can become very large where the original slope is small. Geometrically, reflecting a nearly horizontal tangent across y=x makes a steep tangent. The reciprocal formula requires a nonzero original derivative; a zero slope calls for separate analysis rather than division by zero.