﻿ ﻿DERIVATIVE OF THE POWER FUNCTION - Differentiation of Algebraic Functions - The Calculus Primer

## The Calculus Primer (2011)

### Chapter 9. DERIVATIVE OF THE POWER FUNCTION

3—5. The Derivative of a Variable with a Constant Exponent. Consider the power function y = vn. Applying the General Rule, and expanding by the Binomial Theorem, we obtain:

Step 1.y + Δy = (v + Δv)n.

Step 2.Δy = (v + Δv)nvn. Step 3. Now, as Δx → 0, Δv → 0; hence

Step 4. RULE. The derivative of a variable with a constant exponent is equal to the product of that exponent, the variable raised to a power one less than the original exponent, and the derivative of the variable.

NOTE 1.It is assumed, in the above proof, that the exponent n is a positive integer. If this were not so, we could not have applied the binomial theorem, for when n is negative or fractional, the number of terms in the binomial expansion is infinite; we could not pass to the limit, therefore, since Principle (III) of §1—15 applies only to a finite number of terms. However, we shall find later that the relation holds for any rational value of n, whether positive, negative, or fractional.

NOTE 2.As a special case of formula  above, we let v = x, and obtain: NOTE 3.Formulas  and [5a], in view of what was said in NOTE 1, may be applied to a power function when the exponent is negative or fractional, as shown in Examples 3 and 4 below; thus EXAMPLE 1.Differentiate y = x3 + 5x2 + 4.

Solution. = 3x2 + 10x.

EXAMPLE 2.Differentiate y = 5x4 + 6x3 − 3x2 + 9x − 21.

Solution. = 20x3 + 18x2 − 6x + 9.

EXAMPLE 3.Differentiate y = Solution.y = x−3 + 3x−2 − 5x−1. = −3(x−4) + (3)(−2)(x−3) − 5(−1)(x−2),

or EXAMPLE 4.Differentiate Solution. y = 2x½ + (3x)½ − 4(x−⅓). EXAMPLE 5.Find :y = (5x + 3)2.

Solution.v = 5x + 3, and y = v2.

Hence = nvn−1 = 2v(5) = 10(5x + 3).

This result may be verified by expanding the given function before differentiating; thus

y = (5x + 3)2 = 25x2 + 30x + 9; = 50x + 30 = 10(5x + 3).

EXAMPLE 6.Differentiate with respect to x: y = 4 (x2 + 6x)3.

Solution.

Here v = x2 + 6x, and y = 4v3.

Hence = nvn−1 = (4)(3)v2 = (4)(3)(x2 + 6x)2(2x + 6)

= 12(x2 + 6x)2(2x + 6).

EXAMPLE 7.Differentiate y = Solution.y = (x2 − 4)½;v = x2 − 4 EXERCISE 3—1

Differentiate:

1.y = 2x3 − 5x + 8

2.y = x5 − 3x4 + 2x3 + x

3.y = 4.y = 5 + 6

5.y = v1t + gt2

6.y = x−3 − 3x−2 + 4x−1

7.y = 8.s = 9.y = (x2 − 5x)2

10.s = k(a + 2t)3

11.y = xmmx3

12.y = 13.y = (x3 − 2x)4

14.y = 15.y = 16.y = ﻿