What Is a Derivative?
A derivative measures the instantaneous rate of change of a function — how fast the output is changing at any given input. If you have a function f(x) that describes position over time, the derivative f'(x) tells you the velocity (speed and direction) at each moment. If f(x) describes profit as a function of units sold, f'(x) tells you how much additional profit each additional unit generates.
Geometrically, the derivative at a point equals the slope of the tangent line to the graph at that point. A positive derivative means the function is increasing; a negative derivative means it is decreasing; a zero derivative means the function has a horizontal tangent (potentially a local maximum, minimum, or inflection point).
The formal definition uses a limit: f'(x) = lim(h->0) [f(x+h) - f(x)] / h. This is the limit of the difference quotient — the slope of the secant line as the two points get infinitely close together. While this definition is theoretically important, in practice you will use differentiation rules (shortcuts) that have been derived from this limit definition.
The Power Rule
The power rule is the most frequently used differentiation rule. If f(x) = x^n, then f'(x) = n * x^(n-1). In words: bring the exponent down as a coefficient and reduce the exponent by 1.
Examples: If f(x) = x^5, then f'(x) = 5x^4. If f(x) = x^2, then f'(x) = 2x. If f(x) = x (which is x^1), then f'(x) = 1 (which is x^0 = 1). If f(x) = 7 (a constant, which is 7x^0), then f'(x) = 0.
The power rule works for all real exponents, not just positive integers. If f(x) = x^(-3), then f'(x) = -3x^(-4). If f(x) = x^(1/2) (same as the square root of x), then f'(x) = (1/2)x^(-1/2) = 1 / (2 * sqrt(x)). Rewriting roots and fractions as powers is a crucial skill for applying the power rule.
Combined with the constant multiple rule (the derivative of cf(x) is c * f'(x)) and the sum/difference rule (the derivative of f(x) + g(x) is f'(x) + g'(x)), the power rule lets you differentiate any polynomial. For f(x) = 3x^4 - 7x^2 + 5x - 2, the derivative is f'(x) = 12x^3 - 14x + 5.
The Product Rule
The product rule is used when you need to differentiate a function that is the product of two other functions. If f(x) = u(x) v(x), then f'(x) = u'(x) v(x) + u(x) * v'(x). In words: the derivative of the first times the second, plus the first times the derivative of the second.
Example: Find the derivative of f(x) = x^2 sin(x). Let u = x^2 and v = sin(x). Then u' = 2x and v' = cos(x). By the product rule: f'(x) = 2x sin(x) + x^2 * cos(x).
Example: Differentiate f(x) = (3x + 1)(x^2 - 4). Let u = 3x + 1 (u' = 3) and v = x^2 - 4 (v' = 2x). Product rule: f'(x) = 3(x^2 - 4) + (3x + 1)(2x) = 3x^2 - 12 + 6x^2 + 2x = 9x^2 + 2x - 12. You can verify this by first expanding f(x) = 3x^3 + x^2 - 12x - 4 and differentiating directly: f'(x) = 9x^2 + 2x - 12. Same answer.
A common mistake is thinking the derivative of a product is just the product of the derivatives. It is not. d/dx [u v] is NOT u' v'. You must use the product rule formula. This error is one of the most frequent in calculus courses.
The Quotient Rule
The quotient rule is used when differentiating a function that is one function divided by another. If f(x) = u(x) / v(x), then f'(x) = [u'(x) v(x) - u(x) v'(x)] / [v(x)]^2. A common mnemonic is "low d-high minus high d-low, over low squared" — where "high" is the numerator and "low" is the denominator.
Example: Find the derivative of f(x) = (x^2 + 1) / (x - 3). Let u = x^2 + 1 (u' = 2x) and v = x - 3 (v' = 1). Quotient rule: f'(x) = [2x(x - 3) - (x^2 + 1)(1)] / (x - 3)^2 = [2x^2 - 6x - x^2 - 1] / (x - 3)^2 = (x^2 - 6x - 1) / (x - 3)^2.
Example: Differentiate f(x) = sin(x) / x. Let u = sin(x) (u' = cos(x)) and v = x (v' = 1). Quotient rule: f'(x) = [cos(x) x - sin(x) 1] / x^2 = [x * cos(x) - sin(x)] / x^2.
Many quotient rule problems can also be solved using the product rule by rewriting the quotient as a product with a negative exponent. For example, sin(x)/x = sin(x) * x^(-1), then use the product rule. Some students find this approach simpler. Choose whichever method feels more natural for each problem.
The Chain Rule
The chain rule is used to differentiate composite functions — functions inside other functions. If f(x) = g(h(x)), then f'(x) = g'(h(x)) * h'(x). In words: take the derivative of the outer function (leaving the inner function alone), then multiply by the derivative of the inner function.
Example: Find the derivative of f(x) = (3x + 2)^5. The outer function is u^5 and the inner function is u = 3x + 2. Derivative of outer: 5u^4 = 5(3x + 2)^4. Derivative of inner: 3. Chain rule: f'(x) = 5(3x + 2)^4 * 3 = 15(3x + 2)^4.
Example: Differentiate f(x) = sin(x^2). Outer function: sin(u), derivative: cos(u). Inner function: u = x^2, derivative: 2x. Chain rule: f'(x) = cos(x^2) 2x = 2x cos(x^2).
The chain rule often combines with other rules. For f(x) = e^(3x) cos(2x), you need both the product rule and the chain rule. Let u = e^(3x) (derivative: 3e^(3x) by chain rule) and v = cos(2x) (derivative: -2sin(2x) by chain rule). Product rule: f'(x) = 3e^(3x) cos(2x) + e^(3x) * (-2sin(2x)) = e^(3x)[3cos(2x) - 2sin(2x)].
Derivatives of Trigonometric Functions
The six trigonometric functions have standard derivatives that you should memorize. d/dx[sin(x)] = cos(x). d/dx[cos(x)] = -sin(x). d/dx[tan(x)] = sec^2(x). d/dx[cot(x)] = -csc^2(x). d/dx[sec(x)] = sec(x)tan(x). d/dx[csc(x)] = -csc(x)cot(x).
Notice the pattern: the "co" functions (cosine, cotangent, cosecant) all have negative derivatives. This pattern makes the derivatives easier to remember — if the function starts with "co," the derivative has a negative sign.
When trigonometric functions are composed with other functions, apply the chain rule. d/dx[sin(3x)] = cos(3x) 3 = 3cos(3x). d/dx[tan(x^2)] = sec^2(x^2) 2x. d/dx[cos^2(x)] = d/dx[(cos(x))^2] = 2cos(x) * (-sin(x)) = -2sin(x)cos(x) = -sin(2x).
For inverse trigonometric functions: d/dx[arcsin(x)] = 1/sqrt(1-x^2). d/dx[arccos(x)] = -1/sqrt(1-x^2). d/dx[arctan(x)] = 1/(1+x^2). These appear frequently in integration problems and are worth knowing.
Derivatives of Exponential and Logarithmic Functions
The natural exponential function has the remarkable property that it is its own derivative: d/dx[e^x] = e^x. For a general exponential e^(u), the chain rule gives d/dx[e^u] = e^u u'. For a^x (where a is any positive constant), d/dx[a^x] = a^x ln(a).
The natural logarithm has derivative: d/dx[ln(x)] = 1/x. For ln(u) where u is a function of x: d/dx[ln(u)] = u'/u. For log base a: d/dx[log_a(x)] = 1/(x * ln(a)).
Example: Find d/dx[e^(5x^2)]. Outer: e^u, derivative e^u. Inner: u = 5x^2, derivative 10x. Chain rule: 10x * e^(5x^2).
Example: Find d/dx[ln(x^2 + 1)]. Using d/dx[ln(u)] = u'/u: derivative = 2x / (x^2 + 1). Logarithmic differentiation is a powerful technique for complex products and quotients — take the natural log of both sides, differentiate using log properties, then solve for y'. If you need help differentiating a specific function, ScanSolve can show you the exact rule to apply and walk through each step.
Practice Problems
Problem 1: Find f'(x) if f(x) = 4x^3 - 2x^2 + 7x - 9. Solution: Power rule on each term: f'(x) = 12x^2 - 4x + 7.
Problem 2: Find f'(x) if f(x) = x^2 e^x. Solution: Product rule. u = x^2, u' = 2x, v = e^x, v' = e^x. f'(x) = 2x e^x + x^2 e^x = e^x(2x + x^2) = x e^x(2 + x).
Problem 3: Find f'(x) if f(x) = (2x + 1)^4. Solution: Chain rule. Outer: u^4, derivative 4u^3. Inner: 2x + 1, derivative 2. f'(x) = 4(2x + 1)^3 * 2 = 8(2x + 1)^3.
Problem 4: Find f'(x) if f(x) = sin(x) / (x^2 + 1). Solution: Quotient rule. u = sin(x), u' = cos(x), v = x^2 + 1, v' = 2x. f'(x) = [cos(x)(x^2 + 1) - sin(x)(2x)] / (x^2 + 1)^2.
