25 Summary of derivative rules

sin3

f(x) = x 1+x, D = {x x 6= −1}; g(x) = sin3x, D = R (a) (f g)(x) = f(g(x)) = f(sin3x) = sin3x 1+sin3x Domain: 1+sin3x 6= 0 ⇒ sin3x 6= −1 ⇒ 3x 6=

25 Summary of derivative rules - Auburn University

sin3x 4 (b) f(x) = 4 sin3x Solution When either the numerator or the denominator is constant, the quo-tient rule can be avoided by rst rewriting, and such a solution is generally easier than one using the quotient rule (If neither numerator nor denominator is constant, then rewriting in order to use the product rule requires more steps

CHAPTER 4 FOURIER SERIES AND INTEGRALS

sin3x 3 + sin5x 5 + sin7x 7 +··· Take the derivative of every term to produce cosines in the up-down delta function: Up-down series UD(x)= 4 π [cosx+cos3x+cos5x+cos7x+···] (14) Those coeﬃcients don’t decay at all The terms in the series don’t approach zero, so oﬃcially the series cannot converge Nevertheless it is somehow

Math 115 HW Solutions

0x sin3x = C 1 cos3x+C 2 sin3x Now, we would like to guess that the particular solution is Acos3x+Bsin3x, but both cos3x and sin3x are solutions to the homogeneous equation Hence, we multiply by x and guess that y p = Axcos3x+Bxsin3x Then y0 p = Acos3x−3Axsin3x+Bsin3x+3Bxcos3x = (A+3Bx)cos3x+(B −3Ax)sin3x and so y00