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 = 12 xy y ± ( x 2 + e y ) = 12 xy y ± = 12 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...
View
Full Document
 Spring '07
 JonathanRogawski
 Math, Calculus, Derivative, Implicit Differentiation, Power Rule

Click to edit the document details