Derivatives of Sums, Powers, and Polynomials

Derivatives of Sums, Powers, and Polynomials Taking the derivative of a function essentially boils down to taking a special limit involving that function. Just as we had a rule that allowed us to find the limit of the sum of functions, we have an analogous result for derivatives. Sum Rule for Derivatives df dg d (f (x) + g(x)) = + = f 0 (x) + g 0 (x) dx dx dx Thus, the derivative of a sum is just the sum of the derivatives. Example 1 Find the derivative of 9.8x2 + 5x. Solution Since we have already calculated the derivatives of the individuals terms, we can simply apply the sum rule for derivatives. The derivative of the first term is 19.6x and the second is a linear function so its derivative is 5. Thus, the derivative of the sum is 19.6x + 5 Our next result is very useful; it will lessen the arduous task of calculating many derivatives. Power Rule for Derivatives d p x = pxp−1 dx For any p ∈ R (p is a real number). Example 2 Find the derivative of x2 . Solution We apply the power rule to find d 2 x = 2x2−1 = 2x dx √ Example 3 Find the derivative of x. √ Solution First we rewrite x = x−1/2 and then apply the power rule. d 1/2 1 1/2−1 1 −1/2 1 x = x = x = √ dx 2 2 2 x Since the sqare root function is not real valued for x < 0, the function is not defined for x < 0, and its derivative is not defined for x ≤ 0, as division by 0 is forbidden. Example 4 Find the derivative of x−1 . Solution Since x−1 = x1 is not defined for x = 0, neither is its derivative. Nevertheless, we can apply the power rule to find the value of the derivative at all other points. d −1 x = −x−2 dx Example 5 Find the derivative of x−3/4 . Solution This function, and thus its derivative, is not defined for x ≤ 0. Apply the power rule to find the derivative for other points. 3 d −3/4 x = − x−7/4 dx 4 Example 6 For a general object, surface area and volume are related by S = cV 2/3 where c is a constant of proportionality, which depends on the given object. Apply the power rule to find the derivative, and describe how the proportionality between surface area and volume changes with changing volume. Solution First let us apply the power rule to find that 2 dS = cV −1/3 dV 3 1 . Thus, as volume increases, the The rate at which the surface area changes is proportionate to √ 3 V surface area increases more and more slowly; as an object gets large, the ratio of surface area to volume decreases. With the addition of one more result, we will be able to efficiently calculate the derivatives of arbitrary polynomials. Just as the limit of a constant times a function was the constant times the limit of the function, the same holds true for derivatives. Constant Product Rule for Derivatives If c is a constant and f (x) is a differentiable function we find that d df [cf (x)] = c dx dx Thus, when we are dealing with polynomials, we can use the sum rule to add the derivatives of each term, and the constant product and power rules to calculate the derivative of each individual term. Consider the following example. Example 7 Find the derivative of ax3 + bx2 + cx + d, where a, b, c, d ∈ R (are real numbers). Solution First we apply the sum rule, then the constant product rule, and finally the power rule to find that d 3 d d d dx3 dx2 dx d (ax3 + bx2 + cx + d) = ax + bx2 + cx + d = a +b + c + d = 3ax2 + 2bx + c dx dx dx dx dx dx dx dx In some situations, we may wish to find the derivative of polynomial like (x + 2)4 . Multiplying through all of these terms is a very long task. Fortunately, we can use the binomial theorem to greatly simplify the task. The Binomial Theorem The expansion of (x + y)n is a polynomial of degree n, and (x + y)n = xn + nxn−1 y + n! n! xn−2 y 2 + . . . + xn−k y k + . . . + nxy n−1 + y n (n − 2)!2 (n − k)!k! The above equation is extremely daunting, so we will write the expansion for the first few values of n. (x + y)2 = x2 + 2xy + y 2 (x + y)3 = x3 + 3x2 y + 3xy 2 + y 3 (x + y)4 = x4 + 4x3 y + 6x2 y 2 + 4xy 3 + y 4 The first result should be very familiar, but the others are unlikely so. Example 8 Find the derivative of (2x + 1)3 . Solution One strategy would to be expand the polynomial by hand, but we can use the binomial theorem to find that (2x + 1)3 = (2x)3 + 3(2x)2 + 3(2x) + 1 = 8x3 + 12x2 + 6x + 1 Now we apply our results from earlier to calculate the derivative. d d (2x + 1)3 = (8x3 + 12x2 + 6x + 1) = 24x2 + 24x + 6 dx dx