121616949-math.100 - yx 2 e y = x We treat y as an...

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86 Chapter 4 Transcendental Functions We might have recognized at the start that (1 , - 3) is on the function y = L ( x ) = - 4 - x 2 . We could then take the derivative of L ( x ), using the power rule and the chain rule, to get L ( x ) = - 1 2 (4 - x 2 ) - 1 / 2 ( - 2 x ) = x 4 - x 2 . Then we could compute L (1) = 1 / 3 by substituting x = 1. Alternately, we could realize that the point is on L ( x ), but use the fact that y = - x/y . Since the point is on L ( x ) we can replace y by L ( x ) to get y = - x L ( x ) = - x 4 - x 2 , without computing the derivative of L ( x ) explicitly. Then we substitute x = 1 and get the same answer as before. In the case of the circle it is possible to find the functions U ( x ) and L ( x ) explicitly, but there are potential advantages to using implicit differentiation anyway. In some cases it is more difficult or impossible to find an explicit formula for y and implicit differentiation is the only way to find the derivative. EXAMPLE 4.17 Find the derivative of any function defined implicitly by
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Unformatted text preview: yx 2 + e y = x . We treat y as an unspecified function and use the chain rule: d dx ( yx 2 + e y ) = d dx x ( y · 2 x + y ± · x 2 ) + y ± e y = 1 y ± x 2 + y ± e y = 1-2 xy y ± ( x 2 + e y ) = 1-2 xy y ± = 1-2 xy x 2 + e y You might think that the step in which we solve for y ± could sometimes be difficult— after all, we’re using implicit differentiation here because we can’t solve the equation yx 2 + e y = x for y , so maybe after taking the derivative we get something that is hard to solve for y ± . In fact, this never happens. All occurrences y ± come from applying the chain rule, and whenever the chain rule is used it deposits a single y ± multiplied by some other expression. So it will always be possible to group the terms containing y ± together and...
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