The video explains **implicit differentiation**—the process of differentiating both sides of an equation that relates x and y without solving for y explicitly. By treating dx and dy as infinitesimal changes, the derivative of a expression like s = x² + y² is ds = 2x dx + 2y dy.
- **Circle example**: For x² + y² = 25, differentiating gives 2x dx + 2y dy = 0, so dy/dx = −x/y. At (3, 4) the tangent slope is −3/4.
- **Related‑rates ladder**: Writing x(t)² + y(t)² = 25 and differentiating with respect to time yields 2x x′ + 2y y′ = 0. With y′ = −1 m/s, x = 3, y = 4, we find x′ = 4/3 m/s.
- **Interpretation**: ds = 2x dx + 2y dy measures how s changes for any tiny step (dx, dy). Requiring ds = 0 (no change in s) picks out steps that stay on the level curve; for infinitesimal steps this approximates the tangent line.
- **Other examples**:
- For sin x·y² = x, differentiating gives cos x·y² dx + 2 sin x·y dy = dx, which can be solved for dy/dx.
- To differentiate ln x, rewrite y = ln x as eʸ = x. Differentiating both sides yields eʸ dy = dx, so dy/dx = 1/eʸ = 1/x.
The method shows how tiny nudges in the variables relate, a core idea that extends to multivariable calculus. The video ends by previewing a discussion of limits and the formal definition of the derivative.
1. A circle with radius 5 centered at the origin is described by the equation \(x^2 + y^2 = 5^2\).
2. Every point on this circle is exactly 5 units from the origin, as given by the Pythagorean theorem.
3. The slope of the tangent line to the circle at the point \((3,4)\) is \(-\frac{x}{y} = -\frac{3}{4}\).
4. Implicit differentiation of \(x^2 + y^2 = 25\) yields \(2x\,dx + 2y\,dy = 0\).
5. Solving \(2x\,dx + 2y\,dy = 0\) for \(\frac{dy}{dx}\) gives \(\frac{dy}{dx} = -\frac{x}{y}\).
6. For a 5‑meter ladder leaning against a wall, if the top is 4 m high, the bottom is 3 m from the wall (by the Pythagorean theorem).
7. If the top of the ladder drops at \(1\) m/s (\(\frac{dy}{dt} = -1\) m/s), differentiating \(x(t)^2 + y(t)^2 = 25\) gives \(2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0\).
8. Substituting \(x=3\), \(y=4\), \(\frac{dy}{dt}=-1\) yields \(\frac{dx}{dt} = \frac{4}{3}\) m/s, the speed at which the bottom moves away from the wall.
9. Defining \(s = x^2 + y^2\), the differential \(ds = 2x\,dx + 2y\,dy\) measures the change in \(s\) for tiny steps \(dx, dy\).
10. To remain on the circle, \(s\) must stay constant, so \(ds = 0\) leads to the condition \(2x\,dx + 2y\,dy = 0\).
11. For the implicit curve \(\sin(x)\,y^2 = x\), differentiating both sides gives \(\sin(x)\cdot2y\,dy + y^2\cos(x)\,dx = dx\).
12. Rearranging the previous result provides an expression for \(\frac{dy}{dx}\) on that curve.
13. The natural logarithm curve \(y = \ln x\) can be rewritten as \(e^y = x\).
14. Differentiating \(e^y = x\) yields \(e^y\,dy = dx\), so \(\frac{dy}{dx} = \frac{1}{e^y} = \frac{1}{x}\).
15. Hence the derivative of \(\ln x\) is \(\frac{1}{x}\).
16. Implicit differentiation is a tool used in multivariable calculus to relate tiny changes in multiple variables.