Power Rule of Differentiation

By the power rule (x^n)′ = n·x^(n−1). So you bring the exponent down as a factor and reduce the exponent by one. Both steps belong together. Example: for f(x) = 3x⁴, f′(x) = 3·4·x³ = 12x³; the constant prefactor 3 is kept. For g(x) = x⁻² you get g′(x) = −2·x⁻³. The rule also holds for negative and fractional exponents. A common mistake is to perform only one of the two steps, that is to bring the exponent down but not reduce it, or vice versa. Always watch both steps at the same time.

First write the root as a power with a fractional exponent, then you can apply the power rule directly. The square root is √x = x^(1/2). Its derivative is (1/2)·x^(−1/2) = 1/(2√x). For the cube root ∛x = x^(1/3) with derivative (1/3)·x^(−2/3). The trick is always to rewrite the root as a power, apply the rule and, if needed, represent the result again as a root. This way you avoid special rules and treat roots like any other power function. Watch the negative exponents in the result, which mean a root in the denominator.

When differentiating you reduce the exponent by one and multiply by the old exponent: (x^n)′ = n·x^(n−1). When integrating you do exactly the opposite: you raise the exponent by one and divide by the new exponent, ∫x^n dx = x^(n+1)/(n+1) + C. Integration is the inverse of differentiation, so the rule is mirrored as well. Important is the exception n = −1: for 1/x the integration rule fails, because you would divide by zero; here the integral is the natural logarithm ln|x|. When integrating, also never forget the integration constant C, since the derivative of a constant is zero and therefore remains undetermined.

The power rule (x^n)′ = n·x^(n−1) holds not only for natural numbers but for arbitrary real exponents, so also for negative ones. A negative exponent means a fraction, since x⁻ⁿ is the same as 1/xⁿ. Instead of applying the quotient rule, you conveniently write the fraction as a power with a negative exponent and differentiate directly. Example: 1/x = x⁻¹ has the derivative −1·x⁻² = −1/x². For 1/x² = x⁻² you get −2·x⁻³. The procedure is always the same: bring the exponent down, reduce the exponent by one. Note that a negative exponent reduced by one becomes even more negative, so −2 becomes −3.

You differentiate a polynomial term by term, because the derivative of a sum is the sum of the derivatives. To each individual term you apply the power rule and keep constant prefactors. A constant summand drops out, since the derivative of a constant is zero. Example: for f(x) = 4x³ − 5x² + 2x − 7 you get f′(x) = 12x² − 10x + 2. The linear term 2x becomes 2, because x¹ becomes x⁰ = 1, and the constant −7 vanishes. This term-by-term approach is the basis of every curve discussion, because it lets you quickly determine the first and second derivative of a polynomial.