Page:Elements of the Differential and Integral Calculus - Granville - Revised.djvu/223

 But from (A), $$f(x, y) = 0. \therefore u = 0$$ and $$\tfrac{du}{dx} = 0;$$ that is,

Solving for $$\tfrac{dy}{dx},$$ we get

a formula for differentiating implicit functions. This formula in the form (C) is equivalent to the process employed in &sect;62, for differentiating implicit functions, and all the examples at the end of &sect;63 may be solved by using formula (57). Since

for all admissible values of $$x$$ and $$y$$, we may say that (57) gives the relative time rates of change of $$x$$ and $$y$$ which keep $$f(x, y)$$ from changing at all. Geometrically this means that the point $$(x, y)$$ must move on the curve whose equation is (D), and (57) determines the direction of its motion at any instant. Since

we may write (57) in the form of

Given $$x^2 y^4 + \sin y = 0$$, find $$\tfrac{dy}{dx}$$.


 * Solution. Let $$f(x, y) = x^2 y^4 + \sin y.$$

If $$x$$ increases at the rate of 2 inches per second as it passes through the value $$x = 3$$ inches, at what rate must $$y$$ change when $$y = 1$$ inch, in order that the function $$2xy^2 - 3x^2 y$$ shall remain constant?


 * Solution. Let $$f(x, y) = 2xy^2 - 3x^2 y$$; then


 * Substituting in (57a),


 * But x = 3, y = 1, $$\tfrac{dx}{dt} = 2.$$ &there4; $$\tfrac{dy}{dt} = -2\tfrac{2}{15},$$ ft. per second. Ans.