freeCodeCamp/guide/english/mathematics/implicit-differentiation/index.md

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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 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. 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 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.