tailieunhanh - Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 57

Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 57. A major complaint of professors teaching calculus is that students don't have the appropriate background to work through the calculus course successfully. This text is targeted directly at this underprepared audience. This is a single-variable (2-semester) calculus text that incorporates a conceptual re-introduction to key precalculus ideas throughout the exposition as appropriate. This is the ideal resource for those schools dealing with poorly prepared students or for schools introducing a slower paced, integrated precalculus/calculus course | Implicit Differentiation 541 6. Suppose y 4 where x and g x are positive for all x. Use logarithmic differ- x entiation to flnd Verify that this is the same result you would get had you used the Quotient Rule. 7. If you felt so inclined you could come up with a rule for taking the derivative of functions of the form x Vx where x is positive. You might call it the Tower Rule since you have a tower of functions or you might think of a more descriptive name. In any case what would this rule be EXAMPLE IMPLICIT DIFFERENTIATION When using the process of logarithmic differentiation we differentiate an equation in which y is not explicitly expressed as a function of x. Logarithmic differentiation is a special case of the broader concept of implicit differentiation a concept with far-reaching implications and applications. The basic idea is that we can flnd even when y is not given explicitly as a function of x. We differentiate both sides of the equation that relates x and y applying the Chain Rule to differentiate terms involving y because y varies with x. Implicit differentiation is an important concept we ll begin with a very straightforward example to illustrate what is going on. Consider the circle of radius 2 centered at the It is given by x2 y2 4. Find the slope of the line tangent to the circle at the following points. a 1 V3 b 1 -V3 SOLUTION Although y is not a function of x it can be expressed as two different functions of x. y 4 x2 the top semicircle and y V4 x2 the bottom semicircle Each of these functions gives y explicitly as a function of x. One approach is to differentiate the former expression to get information about the point 1 V3 and the latter to deal with 5This circle is the set of all points a distance 2 from the origin. If x y is a point on this circle then the distance formula tells us that y x 0 2 y 0 2 2. And conversely if x y satisfles the equation x2 y2 2 then x y is a point on the circle. Therefore x2 y2 4is the .