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Implicit differentiation

12 November, 2015 - 11:22

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 be- tween 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 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 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 re- ally just 2dx, and we’ve really just computed the derivative of x^2with respect to x and multiplied it by dx. In symbols,

\begin{align*} d(x^2) &=\frac{d(x^2)}{dx}\cdot dx \\ &= 2xdx \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 use- less. For instance, if we’re given a numerical value of x, we can al- ways use Newton’s method to find y, and then evaluate the derivative.