Implicit Differentiation Khan Academy

How Implicit Differentiation Actually Works

You start with an equation where y isn't isolated, like x squared plus y squared equals 25. Most students panic here because they're trained to solve for y first. Don't. Differentiate both sides with respect to x anyway, and whenever you hit a y term, multiply by dy/dx because y is secretly a function of x. That's the whole trick. The derivative of y squared is 2y times dy/dx, not just 2y. You'll see this mistake constantly in homework submissions. Khan Academy breaks this down across a handful of short videos, starting from the basic concept and moving into increasingly complex examples. The exercise sets are paired with each video, which is useful if you actually work through them instead of just watching. I used their materials while tutoring calculus students and found them solid for building initial intuition, though they don't push hard enough on the edge cases where things go wrong.

Implicit Differentiation Khan Academy walkthrough

The Khan Academy section on implicit differentiation is organized into a few logical modules. You begin with differentiating both sides of an equation, then move into finding derivatives of functions defined implicitly, and eventually tackle related rates applications that depend on the technique. Each lesson runs between three and eight minutes. The practice problems range from straightforward to moderately tricky, and the step-by-step hints are genuinely helpful when you get stuck rather than just giving away the answer. One thing the videos don't emphasize enough: when you differentiate, you're not solving for y first and then differentiating. You're treating the entire equation as a balance and applying the derivative operator to every term equally. This distinction matters when you hit something like x times y equals 1, where implicit differentiation immediately gives you y plus x times dy/dx equals 0, and solving for dy/dx takes one extra algebra step. If you tried to solve for y first here, you'd get y equals 1 over x and then differentiate using the quotient rule, which works but defeats the purpose of learning the technique. The point is to build the habit of not isolating variables before differentiating. I ran into a specific problem a few years ago while working through a problem set that involved a circle equation offset from the origin: x minus 2 squared plus y plus 3 squared equals 16. The standard approach works fine, but I kept getting sign errors when applying the chain rule to the y term because the inner derivative of y plus 3 with respect to x is just dy/dx, and the negative sign from differentiating y plus 3 got lost in my notes. The workaround was to label every d/dx operation explicitly on the first pass and never skip a step, even the trivial ones. Writing out d/dx of y plus 3 equals dy/dx on the page made the error impossible to miss. It added maybe thirty seconds per problem but caught mistakes that would have cost ten minutes of debugging later.

Here's a counter-intuitive point that beginners usually miss: implicit differentiation doesn't always require you to solve for dy/dx at the end. Sometimes the problem asks for the derivative at a specific point, and you can substitute the coordinates directly into the differentiated equation before isolating dy/dx. This is faster and reduces algebra errors significantly. For example, if you have x cubed plus y cubed equals 6xy and you need the derivative at the point where x equals 2 and y equals 2, you substitute those values right after differentiating rather than simplifying the entire expression for dy/dx first. The algebra is cleaner and the numerical answer comes out quicker. Another thing worth noting: implicit differentiation fails or becomes meaningless when the equation defines multiple branches of a relation and you need to track which branch you're on. Take x squared minus y squared equals 1. Differentiating gives you 2x minus 2y times dy/dx equals 0, so dy/dx equals x over y. But this derivative is undefined at y equals 0, which corresponds to the vertices of the hyperbola at positive and negative 1 on the x-axis. The formula doesn't lie, but students sometimes treat the derivative as valid everywhere and miss that vertical tangents exist at those points. Khan Academy covers this indirectly through examples but doesn't make the limitation explicit enough. The bigger limitation of the Khan Academy materials is that the practice problems tend to stay in the comfort zone. You'll see circles, ellipses, and occasionally a cubic curve. You won't see much in the way of transcendental equations mixed with implicit forms, like e to the y plus x times y equals 1, which requires combining implicit differentiation with logarithmic differentiation concepts. If you're preparing for a rigorous exam or AP Calculus BC, you'll want supplemental problems from a textbook or a different resource that pushes further.

For the actual technique, here's the practical sequence I recommend. Write down the original equation. Apply d/dx to every single term on both sides. For any term containing only x, differentiate normally. For any term containing y, differentiate as if y were x and then multiply by dy/dx. For products or quotients involving both x and y, use the product or quotient rule and remember the chain rule component on every y factor. After differentiating, collect all dy/dx terms on one side and everything else on the opposite side. Factor out dy/dx. Divide to isolate it. Simplify if needed. That's it. Thirty seconds to two minutes per problem depending on complexity. If you're just starting out, spend about twenty minutes on the first two Khan Academy videos and then do the accompanying exercises until you score above seventy percent without hints. Move on to the harder modules only after that baseline is solid. The total time investment for a functional grasp of implicit differentiation through their platform is roughly two to three hours including practice, and it covers about eighty percent of what you'd encounter in a standard calculus course.