Differentiation & Derivatives Calculator

Differentiation Guide

What do I need to know about The Power Rule?

For f(x) = axⁿ: f'(x) = anx^(n-1). Examples: d/dx(3x⁴) = 12x³. d/dx(5x) = 5. d/dx(7) = 0 (constants differentiate to zero). d/dx(x^(1/2)) = (1/2)x^(-1/2) = 1/(2√x). d/dx(1/x) = d/dx(x^(-1)) = -x^(-2) = -1/x². Polynomials: differentiate each term separately. d/dx(3x⁴ − 2x² + 5x − 7) = 12x³ − 4x + 5. Chain rule: d/dx(f(g(x))) = f'(g(x)) × g'(x). Example: d/dx((3x+1)⁵) = 5(3x+1)⁴ × 3 = 15(3x+1)⁴.

What's the key thing to understand about Standard Derivatives?

Trigonometric: d/dx(sin x) = cos x. d/dx(cos x) = −sin x. d/dx(tan x) = sec²x. With chain rule: d/dx(sin(ax)) = a cos(ax). d/dx(cos(ax)) = −a sin(ax). Exponential: d/dx(eˣ) = eˣ (unique — its own derivative). d/dx(e^(ax)) = a e^(ax). d/dx(aˣ) = aˣ ln a. Logarithmic: d/dx(ln x) = 1/x. d/dx(ln(ax)) = 1/x. d/dx(log_a(x)) = 1/(x ln a). These standard results are given in examination data sheets — the key is knowing which rule to apply.

What should I know about Finding Stationary Points?

Stationary points: where dy/dx = 0. To find them: differentiate and set equal to zero. Solve for x. Substitute back to find y coordinates. Nature: use second derivative test. If d²y/dx² > 0: minimum. If d²y/dx² < 0: maximum. If d²y/dx² = 0: inconclusive (use first derivative test — check sign of dy/dx either side of stationary point). Example: y = x³ − 6x² + 9x. dy/dx = 3x² − 12x + 9 = 3(x−1)(x−3) = 0. x = 1 or x = 3. d²y/dx² = 6x − 12. At x=1: −6 < 0 → maximum. At x=3: 6 > 0 → minimum.

What do I need to know about Product and Quotient Rules?

Product rule: d/dx(uv) = u'v + uv'. Example: d/dx(x² sin x) = 2x sin x + x² cos x. Quotient rule: d/dx(u/v) = (u'v − uv') / v². Example: d/dx(sin x / x) = (cos x · x − sin x · 1) / x² = (x cos x − sin x) / x². Implicit differentiation: for equations of the form f(x,y) = 0. Differentiate both sides with respect to x. Any y terms gain a factor dy/dx. Useful for finding gradients of circles and other implicitly defined curves.

What's the key thing to understand about the Chain Rule?

The chain rule differentiates composite functions — a function nested inside another function, such as (3x + 1)⁵, where the inner function (3x+1) is raised to a power by the outer function. The rule states: if y = f(g(x)), then dy/dx = f'(g(x)) × g'(x) — differentiate the outer function (keeping the inner function unchanged inside it), then multiply by the derivative of the inner function. For example, to differentiate (3x+1)⁵: the outer function is (something)⁵ with derivative 5(something)⁴, and the inner function is 3x+1 with derivative 3, so the chain rule gives d/dx[(3x+1)⁵] = 5(3x+1)⁴ × 3 = 15(3x+1)⁴. The chain rule is essential because most real functions encountered in calculus are composites rather than simple single-step expressions, and forgetting to multiply by the inner function's derivative is one of the most common errors students make — it's the rule most responsible for turning a seemingly complex derivative into a manageable, mechanical calculation.

What should I know about Second Derivatives and Concavity?

The second derivative, f''(x), is simply the derivative of the derivative — it measures how the rate of change itself is changing, which translates geometrically into the concavity (curvature) of a function's graph. When f''(x) is positive over an interval, the graph is concave up (curving like the inside of a bowl), and a stationary point found in that interval (where f'(x) = 0) is a local minimum. When f''(x) is negative, the graph is concave down (curving like a dome), and a stationary point there is a local maximum. This test — called the second derivative test — provides a faster alternative to checking the sign of the first derivative on either side of a stationary point. For example, for f(x) = x³ − 3x, the stationary points occur at x = ±1; since f''(x) = 6x, f''(1) = 6 (positive, a local minimum) and f''(−1) = −6 (negative, a local maximum). A second derivative of exactly zero at a stationary point is inconclusive and signals a possible inflection point, requiring further investigation.

What do I need to know about Implicit Differentiation?

Implicit differentiation is used when y is not isolated as an explicit function of x — for example, in the equation of a circle, x² + y² = 25, y isn't written as "y = ..." directly. To differentiate such an equation, differentiate both sides with respect to x, treating y as a function of x and applying the chain rule to any y term (so d/dx(y²) becomes 2y·dy/dx, not just 2y), then solve the resulting equation for dy/dx. For x² + y² = 25, differentiating gives 2x + 2y(dy/dx) = 0, so dy/dx = -x/y. This technique is essential whenever a curve's equation can't be easily rearranged into y = f(x) form, which includes circles, ellipses, and many other implicitly defined curves that come up throughout calculus and related rates problems.

What should I know about Differentiation from First Principles?

Every differentiation rule used in practice — the power rule, the chain rule, the product rule — ultimately derives from a single formal definition: the derivative of f(x) is f'(x) = lim(h→0) [f(x+h) − f(x)]/h, provided this limit exists. This expression captures the derivative's core meaning precisely: it is the limiting value of the gradient of a straight line (a secant) joining two points on the curve, (x, f(x)) and (x+h, f(x+h)), as the gap h between those points shrinks toward zero, so the secant line's gradient approaches the gradient of the tangent line at x. Differentiating from first principles means working directly from this limit definition rather than quoting a memorised rule, and it is how the standard rules are actually proven: for f(x) = x², first-principles differentiation gives f'(x) = lim(h→0) [(x+h)² − x²]/h = lim(h→0) [2xh + h²]/h = lim(h→0) (2x + h) = 2x, confirming the power rule's result for n = 2 directly from the definition. While first-principles differentiation is rarely used for everyday calculations once the standard rules are known, it remains essential for understanding why those rules work, for differentiating unusual functions the standard rules don't directly cover, and for building the rigorous foundation that calculus rests on.

Derivatives of Inverse Trigonometric and Hyperbolic Functions

Beyond the basic power, trig, and exponential rules, these derivatives come up often enough in calculus courses to be worth a dedicated reference:

Notice arcsin and arccos have derivatives that are negatives of each other — which makes sense, since arcsin(x) + arccos(x) = π/2 is a constant, so their derivatives must sum to zero.

Worked Example: Combining the Chain Rule and Product Rule

Differentiate f(x) = (2x + 1)³ cos(x), which requires both the product rule (for the two multiplied factors) and the chain rule (for the cubed binomial). Let u = (2x + 1)³ and v = cos(x). By the product rule, f'(x) = u'v + uv'. For u' (chain rule): if u = (2x+1)³, then u' = 3(2x+1)² × 2 = 6(2x+1)². For v': v' = −sin(x). Combining: f'(x) = 6(2x+1)²cos(x) + (2x+1)³(−sin(x)) = 6(2x+1)²cos(x) − (2x+1)³sin(x). Evaluate at x = 0: (2(0)+1) = 1, so f'(0) = 6(1)²cos(0) − (1)³sin(0) = 6(1)(1) − (1)(0) = 6 − 0 = 6. As a numerical sanity check using the definition of the derivative with a small step h = 0.001: f(0.001) = (1.002)³cos(0.001) ≈ 1.006012 × 0.9999995 ≈ 1.006011, and f(0) = (1)³cos(0) = 1. The slope estimate (f(0.001) − f(0))/0.001 ≈ (1.006011 − 1)/0.001 ≈ 6.011 — closely matching the exact analytical result of 6, confirming the chain-and-product-rule derivative is correct.

Using Derivatives to Find Velocity from a Position Function

One of the most common real-world uses of differentiation is finding instantaneous velocity from a position function: if s(t) describes position over time, then v(t) = s'(t) is the velocity function. A particle's position is given by s(t) = 4t³ − 2t (meters, with t in seconds). Find its velocity at t = 2 seconds. Differentiate using the power rule term by term: s'(t) = 12t² − 2. Evaluate at t = 2: v(2) = 12(2)² − 2 = 12(4) − 2 = 48 − 2 = 46 m/s. You can verify this is reasonable using the definition of the derivative with a small time step, h = 0.001: s(2.001) = 4(2.001)³ − 2(2.001) = 4(8.012006) − 4.002 = 32.048024 − 4.002 = 28.046024, and s(2) = 4(8) − 4 = 32 − 4 = 28. The slope estimate (28.046024 − 28)/0.001 = 46.024 — closely matching the exact value of 46 m/s, confirming the derivative correctly captures the particle's instantaneous rate of change of position, i.e., its velocity, at that exact moment.