### 5.2. Implicit Differentiation

We 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:

y7+y = x7+x2 .

There is no equivalent of the quadratic formula for seventh-order 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 x2 term will grow to (x+dx)2=x+2dx+dx2. The squared infinitesimal is negligible, so the increase in x2 was really just 2dx, and we've really just computed the derivative of x2 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 x2 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.