Implicit Differentiation and Derivatives of Inverse Functions
Differentiate a relation that is not solved for y, and use the same move to get the derivative of an inverse function from the derivative of the original.
What a learner can do afterwards
- Find dy/dx for x² + y² = 25 and read the gradient at a point on the circle
- Derive the derivative of arcsin x from sin y = x
- Explain why dy/dx appears as a factor whenever y is differentiated
1 · Read
Some curves never solve neatly for y, like the circle x squared + y squared = 25. Differentiate both sides with respect to x and treat y as a function of x, so y squared gives 2y times dy/dx. Collecting gives dy/dx = -x/y, so at (3, 4) the gradient is -0.75 and the tangent is y = -0.75x + 6.25.
Inverse derivatives fall out the same way. Write y = arcsin x as sin y = x, differentiate to cos y times dy/dx = 1, and trade cos y for sqrt(1 - x squared) to get 1/sqrt(1 - x squared). Arccos takes the negative twin, arctan x gives 1/(1 + x squared), and arctan(3x) adds a factor 3. In general (f inverse)' = 1 / f'(f inverse(x)), so f(2) = 5 with f'(2) = 3 gives inverse gradient 1/3 at 5.
For xy = 12 at (3, 4), the product rule gives y + x times dy/dx = 0, so dy/dx = -y/x = -4/3. For x squared + xy + y squared = 37 at (3, 4), collecting gives dy/dx = -(2x + y)/(x + 2y) = -10/11, about -0.91.
Differentiate each y term with its dy/dx attached, gather them, divide, then put in the point.
2 · Watch
Take it off screen
Where it sits
8 questions wait behind this lesson, each with its answer explained. Every answer feeds the sky: stars light as they are learned, and dim when it is time to come back.