Math 19: Calculus Winter 2008 Instructor: Jennifer KlokeLecture Outline (Product Rule and QuotientRule)Wednesday, February 6RecapLast time we developed a number of shortcuts for taking the derivative of ex.Today we are going to learn how to take the derivative of products and quotients, whichwill greatly expand our differentiation abilities!!!The Derivative of a ProductNotice: The limit of the product is not the produc t of the limits!:ddx(x ∗ x) =ddx(x2) = 2xbutddx(x) ∗ddx(x) = 1 ∗ 1 6= 2x.Suppose you wanted to differentiate a function that was a product, say f(x)g(x). Forexample, how do you differentiate xex? Using the limit definition of the derivative we canderive:The Product Rule for derivatives is:ddx(f(x)g(x)) = f0(x)g(x) + f(x)g0(x).Mantra: ”The derivative of the product is the derivative of the first times the secondplus the derivative of the second times the first.”Examples1. Let y = xex. Find y0.2. Let y = e2x. Find y0.3. Let z = 2xex(4ex− 3√x). Finddzdx.1The Quotient RuleSuppose we have a function that is a quotient: h(x) =f(x)g(x). What is h0(x)? We can derivethe Quotient Rule from the Product Rule:h(x) =f(x)g(x)h(x)g(x) = f(x)h0(x)g(x) + g0(x)h(x) = f0(x) Take the derivative of both sides and use the product rule on the left.h0(x)g(x) = f0(x) − h(x)g0(x)h0(x) =f0(x) − h(x)g0(x)g(x)=f0(x) −f(x)g(x)g0(x)g(x)substitute h(x) = f(x)/g(x) in=f0(x)g(x) − f(x)g0(x)g(x)2So the Quotient Rule is:ddxf(x)g(x)=f0(x)g(x) − f(x)g0(x)[g(x)]2Notice that in the formula above, it is VERY important that the function in the numeratoris called f(x) and the function in the denomintator is called g(x): the minus sign in ourexpression will give us problems if we switch these names around! Be careful!Examples1. Let y =x2−1x+2. Find y0.2. Let y =xexx+1. Find the equation of the tangent line to this curve at the point (1, e/2).What to Know/Memorize1. The Product Rule:ddx(f(x)g(x)) = f0(x)g(x) + f(x)g0(x).2. The Quotient Rule:ddxf(x)g(x)=f0(x)g(x) −
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