diff --git a/guide/english/mathematics/implicit-differentiation/index.md b/guide/english/mathematics/implicit-differentiation/index.md
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@@ -3,13 +3,30 @@ title: Implicit Differentiation
---
## Implicit Differentiation
-This is a stub. Help our community expand it.
+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.
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+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
-#### More Information:
-
+2x + 2y × y' = 0,
+and so y' = -2x/2y = -x/y.
+For a more involved example, consider the equation exy = e2x - e3y. Taking the derivative of the left hand side gives
+
+exyd(xy)/dx = exy[y + xy'],
+
+using the chain rule and then the product rule. Similarly, the derivative of the right hand side is
+
+2e2x - e3yd(3y)/dx = 2e2x - 3e3yy'
+
+Thus, solving for y', we find
+
+y' = (2x2x - yexy)/(xexy + 3e3y)
+
+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.