--- title: Implicit Differentiation --- ## Implicit Differentiation When differentiating a function, such as f(x) = x², we are differentiating an equation y = x² that has been solved for y in terms of x to find dy/dx. However, there are equations where such a solution is [not practical](https://www.wolframalpha.com/input/?i=solve+y%5E2+%2B+sqrt(y)+%3D+x%5E3+%2B+x+%2B+1+for+y) and some equations simply cannot be solved for y in terms of x, so in order to compute derivatives for these cases we need another approach. This is where *implicit* differentiation comes in, which is roughly just the [chain rule](https://en.wikipedia.org/wiki/Chain_rule). Recall that if we have a function f(x), then the chain rule tells us the derivative of f(x)² is 2f(x) × f'(x). But, despite not having a solution for y in terms of x, this same rule applies. The derivative of y² is 2y × y'. Now we can implicitly differentiate an equation and attempt to solve for y'. For example, the circle of radius 1 is given by the equation x² + y² = 1. Instead of solving for y and taking a derivative of each half of the square root we can simply (implicitly) differentiate both sides with respect to x to find

2x + 2y × y' = 0,

and so y' = -2x/2y = -x/y. For a more involved example, consider the equation e^xy = e^2x - e^3y. Taking the derivative of the left hand side gives

e^xyd(xy)/dx = e^xy[y + xy'],

using the chain rule and then the product rule. Similarly, the derivative of the right hand side is

2e^2x - e^3yd(3y)/dx = 2e^2x - 3e^3yy'

Thus, solving for y', we find

y' = (2x^2x - ye^xy)/(xe^xy + 3e^3y)

In general, if we have an equation f(x,y) = 0, taking the derivative with respect to x gives

∂f/∂x + ∂f/∂y × dy/dx

where we use [partial derivatives](https://en.wikipedia.org/wiki/Partial_derivative) and the chain rule here, so whether one can solve an equation for y or not, it is always possible to get an equation for the derivative y'. Moreover, as the examples above show, it is frequently easier to implicitly differentiate instead of trying to solve for y first and then take the derivative.