5.2. Implicit DifferentiationWe can differentiate any function that is written as a formula, and find a result in terms of a formula. However, sometimes the original problem can't be written in any nice way as a formula. For example, suppose we want to find dy/dx in a case where the relationship between x and y is given by the following equation: y^{7}+y = x^{7}+x^{2} . There is no equivalent of the quadratic formula for seventhorder polynomials, so we have no way to solve for one variable in terms of the other in order to differentiate it. However, we can still find dy/dx in terms of x and y. Suppose we let x grow to x+dx. Then for example the x^{2} term will grow to (x+dx)^{2}=x+2dx+dx^{2}. The squared infinitesimal is negligible, so the increase in x^{2} was really just 2dx, and we've really just computed the derivative of x^{2} with respect to x and multiplied it by dx. In symbols, \begin{align} d(x^2) &= \frac{d(x^2)}{dx} \cdot dx \\ &= 2x dx . \end{align} That is, the change in x^{2} is 2x times the change in x. Doing this to both sides of the original equation, we have \begin{align} d(y^7+y) &= d(x^7+x^2) \\ 7y^6dy + 1dy &= 7x^6dx + 2xdx \\ (7y^6+1)dy &= (7x^6+2x)dx \\ \frac{dy}{dx} &= \frac{7x^6+2x}{7y^6+1} . \end{align} This still doesn't give us a formula for the derivative in terms of x alone, but it's not entirely useless. For instance, if we're given a numerical value of x, we can always use Newton's method to find y, and then evaluate the derivative.
