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Section 3.2 The Product and Quotient Rules. Goals Learn formulas for the derivatives of the product and quotient of two functions whose derivatives are known. The Product Rule. Suppose that f ( x ) and g ( x ) are each known; what is ( fg ) ( x ) ?
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Section 3.2The Product andQuotient Rules • Goals • Learn formulas for the derivatives of the • product and • quotient of two functions whose derivatives are known
The Product Rule • Suppose that f(x) and g(x) are each known; what is (fg) (x) ? • It is tempting to suppose that (fg) (x) = f(x)g(x) • But this is wrong! • Try it with f(x) = x and g(x) = x2 , for example.
Product Rule (cont’d) • To see the correct formula, we first assume that u = f(x) and v = g(x) are both positive differentiable functions. • Then the product uv is an area of a rectangle, as shown on the next slide. • If x changes by an amount ∆x , then the corresponding changes in u and v are
Product Rule (cont’d) • The new value (u + ∆u)(v + ∆v) of the product is the area of the large rectangle in Fig. 1… • if ∆u and ∆v both happen to be positive. • The change in the area of the rectangle is ∆(uv) = (u + ∆u)(v + ∆v) – uv = u∆v + v∆u + ∆u∆v which is the sum of the three shaded areas.
Product Rule (cont’d) • Dividing by ∆x gives • Finally we let ∆x 0 , leading to the calculations shown on the next slide. • Note that ∆u 0 as ∆x 0 since f is differentiable and therefore continuous.
Product Rule (cont’d) • We have assumed that all quantities involved above are positive, however… • …the calculations are valid even if not. • This leads to:
Product Rule • In words, the Product Rule says that • The derivative of a product of two functions is • the first function times the derivative of the second • plus the second function times the derivative of the first.
Example • If f(x) = xex , find • f(x) ; • the nth derivative, f(n)(x) . • Solution We use the Product Rule: • First,
Solution (cont’d) • Then Applying the Product Rule further gives • f(x) = (x + 3)ex • f(4)(x) = (x + 4)ex In general, f(n)(x) = (x + n)ex
Example • If where g(4) = 2 and g(4) = 3 , find f(4) . • Solution The Product Rule gives
The Quotient Rule • We find a rule for differentiating the quotient of two differentiable functionsu = f(x) and v = g(x) in much the same way as for the Product Rule. • If x , u , and v change by amounts ∆x , ∆u , and ∆v , then the corresponding change in the quotient u/v is:
Quotient Rule (cont’d) • Once again, as ∆x 0 , ∆v 0 also, because g is differentiable and therefore continuous. • Thus the Limit Laws give • This leads to the Quotient Rule:
Quotient Rule (cont’d) • So, the derivative of a quotient is • the denominator times the derivative of the numerator • minus the numerator times the derivative of the denominator, • all divided by the square of the denominator.
Example • Find y if • Solution The Quotient Rule gives
Example • Find an equation of the tangent line to the curve y = ex/(1 + x2) at the point (1, e/2) . • Solution According to the Quotient Rule,
Solution (cont’d) • So the slope of the tangent line at (1, e/2) is • Thus the tangent line… • is horizontal, and • its equation is y = e/2 . • The curve and tangent line are graphed on the next slide:
Table of Formulas • Here is a summary of our differentiation formulas so far:
Review • Two ways to find the derivative of combinations of differentiable functions: • The Product Rule • The Quotient Rule