Differentiation Rules Calculating a derivative using the definition can be quite tiresome. Luckily, unless you're specif- ically asked to use the definition, there are a number of derivative rules you can use to calculate derivatives. Here is a list of those rules: d • (xn) = nxn−1. dx d • (c) = 0, where c 2 . dx R d d d • (a · f(x) + b · g(x)) = a · f(x) + b · f(x), where a; b 2 . dx dx dx R d • (ex) = ex dx d • (ax) = ax ln (a) dx d 1 • (ln (x)) = dx x d 1 • (log (x)) = dx b x · ln (b) All of the above derivatives should be learned by all calculus students. The derivatives of the trig functions (below) should be learned by MATH1500 and MATH1510 students, but will not be required of MATH 1520 students. d • (sin (x)) = cos (x) dx d • (cos (x)) = − sin (x) dx d • (tan (x)) = sec2 (x) dx d • (sec (x)) = sec (x) tan (x) dx d • (csc (x)) = − csc (x) cot (x) dx d • (cot (x)) = − csc2 (x) dx 1 Given that f(x) and g(x) are differentiable functions whose derivatives are represented by f 0(x) and g0(x) respectively, the following rules also apply. They must be learned. The Product Rule: d (f(x) · g(x)) = f(x) · g0(x) + g(x) · f 0(x) dx The Quotient Rule: d f(x) g(x) · f(x) − f(x)g0(x) = dx g(x) (g(x))2 The Chain Rule: d (f(g(x))) = f 0(g(x)) · g0(x) dx Example. For each of the following functions, find f 0(x) 1. f(x) = 7x5 − 4x9. f 0(x) = 35x4 − 36x8 p 2. f(x) = 5 x7. p5 7 f(x) = x7 = x 5 0 7 2 =) f (x) = x 5 5 3. f(x) = π4. π4 is a constant, so f 0(x) = 0 . 6 4. f(x) = . x4 6 f(x) = = 6x−4 x4 24 =) f 0(x) = −24x−5 = − x5 2 5. f(x) = log4 (x). 1 f 0(x) = x · ln (4) 6. f(x) = 7x. f 0(x) = 7x · ln (7) 7. f(x) = tan (x) + cot (x). f 0(x) = sec2 (x) + sec (x) tan (x) 8. f(x) = x3 · 3x. Here, we need to use the product rule: f 0(x) = x3 · 3x ln (3) + 3x · 3x2 p 9. f(x) = ln (x) · x. p 1 First, notice that f(x) = ln (x) · x = ln (x) · x 2 . 0 1 − 1 f (x) = ln (x) · x 2 2 log (x) 10. f(x) = 5 . x5 − 3x4 The quotient rule is required for this problem. 1 (x5 − 3x4) · − log (x) · (5x4 − 12x3) x ln (5) 5 f 0(x) = (x5 − 3x4)2 ln (x) 11. f(x) = . ex 1 ex · − ln (x)ex f 0(x) = x (ex)2 3 12. f(x) = (x4 − 3x6)3. The chain rule is required for this problem and the rest. f 0(x) = 3(x4 − 3x6)2 · (4x3 − 18x5) 13. f(x) = ln (x3 + 6x5). 1 f 0(x) = · (3x2 + 30x4) x3 + 6x5 14. f(x) = sin (cos (x2)). Here, the chain rule must be applied twice: f 0(x) = cos (cos (x2)) · (− sin (x2)) · 2x 15. f(x) = px5 ln (x). p 5 1 First, note that f(x) = x5 ln (x) = (x ln (x)) 2 . Here we need to use the chain rule and product rule consecutively. 0 1 5 − 1 5 1 4 f (x) = (x ln (x)) 2 · x · + ln (x) · 5x 2 x log cos (x3 + x) 16. f(x) = 5 . ex5+3x For the problem, both the chain rule and the quotient rule are required. 5 1 5 ex +3x · · (3x2 + 1) − log cos (x3 + x) · ex +3x · (5x4 + 3) (x3 + x) ln (5) 5 f 0(x) = (ex5+3x)2 4 Exercises Differentiate each of the following functions. Note: Some of these functions look like they require the product or quotient rule at a glance, but with a little manipulation, they aren't needed. Try to identify which ones. 3 1. f(x) = x8 4 p 2. f(x) = x − x 3. f(x) = (x3 + 2x)ex p 4. f(x) = ex(x + x x) x2 + 2 5. f(x) = x4 − 3x2 + 1 1 3 6. f(x) = − x + 5x3 x2 x4 p x3 − 2x x 7. f(x) = x 8. f(x) = (3x2 + 2x)7 9. f(x) = sin(4x) (x2 + 8)4 10. f(x) = x + 2 11. f(x) = (x3 + 3)6 sin5(3x2) 5.