Differentiation: the power rule

A constant has derivative zero. The derivative of bx is b. Sums and differences can be differentiated term by term:

f(x) = 3x⁴ − 5x² + 7x − 4
f′(x) = 12x³ − 10x + 7

The constant −4 disappears; the linear term 7x becomes 7. Do not keep an extra x in that last term. A vertical translation changes a curve’s height but not its gradient function.

The n = 0 case is the constant rule. Write its derivative as 0 directly, rather than leaving a misleading expression 0x⁻¹ at x = 0.

Use index laws to turn roots and denominators into powers:

1/x³ = x⁻³
√x = x^1/2
1/√x = x^−1/2
x²/√x = x^3/2, for x > 0

Do not differentiate the numerator and denominator separately and divide their derivatives. That is not a differentiation rule.

For h(x) = (x² − 2)(x + 3), expand first:

h(x) = x³ + 3x² − 2x − 6
h′(x) = 3x² + 6x − 2

Multiplying the two individual derivatives would give 2x, which is wrong. Expanding turns this example into a sum you already know how to differentiate.

For 1/x², both the function and its derivative exclude x = 0. For √x, the function exists at zero but its ordinary finite derivative there does not. For real cube-root powers, negative inputs can be allowed: x^2/3 means (∛x)², with derivative (2/3)x^−1/3 for x ≠ 0.

For a general real exponent, using x > 0 avoids ambiguity. Integer powers extend naturally to negative x, and some rational powers do too. Check the original expression and the point in question.

For f(x) = x³ − 3x + 2, find points where the gradient is 9.

f′(x) = 3x² − 3
3x² − 3 = 9 ⇒ x = ±2
f(2) = 4,   f(−2) = 0

The points are (2,4) and (−2,0). Both roots are needed. If instead a question specifies where a curve meets a line, first solve their intersection equation; then evaluate the derivative at each intersection.

A quadratic f(x) = ax² + bx + c has a stationary point at (2,−1) and passes through (−1,8). Use both coordinates of the stationary point:

f′(2) = 0 ⇒ 4a + b = 0
f(2) = −1 ⇒ 4a + 2b + c = −1
f(−1) = 8 ⇒ a − b + c = 8

From b = −4a and c = 4a − 1, the third equation gives 9a − 1 = 8. Hence a = 1, b = −4, c = 3.

f(x) = 3x⁴ − 5x² + 7x − 4
f′(x) = 12x³ − 10x + 7
f″(x) = 36x² − 10

The second derivative measures how the first derivative changes. It is not the square of dy/dx, and d²y/dx² is not a fraction to cancel mechanically.

The finite binomial expansion of (1 + 2x)⁶ begins 1 + 12x + 60x², followed by terms in x³ and higher. Differentiating the full polynomial gives:

f′(x) = 12 + 120x + terms in x² and higher
For sufficiently small |x|: f′(x) ≈ 12 + 120x

To obtain the derivative through its linear term, retain the original expansion through x². This reasoning works because the omitted terms here are known polynomial powers. An arbitrary numerical approximation cannot automatically be differentiated with the same error guarantee.