3. Implicit Differentiation Solver — Theory
In implicit differentiation, \(y\) is not solved as a function \(y=f(x)\).
Instead, we have an equation relating \(x\) and \(y\), such as
\[
x^2y + y^3 = 5.
\]
We treat \(y\) as a function of \(x\) (so \(y=y(x)\)), and differentiate both sides with respect to \(x\).
1) Key idea
Since \(y\) depends on \(x\), the derivative of \(y\) is not \(0\) but:
\[
\frac{d}{dx}(y) = \frac{dy}{dx}.
\]
More generally, for any function \(g(y)\),
\[
\frac{d}{dx}\big(g(y)\big) = g'(y)\,\frac{dy}{dx}
\]
which is exactly the chain rule.
2) Worked example
Example equation:
\[
x^2y + y^3 = 5
\]
Step A — Differentiate both sides
Differentiate term-by-term. For \(x^2y\) use the product rule:
\[
\frac{d}{dx}(x^2y) = (x^2)'y + x^2(y)' = 2xy + x^2y'
\]
For \(y^3\), use the chain rule:
\[
\frac{d}{dx}(y^3) = 3y^2\,y'
\]
The derivative of the constant \(5\) is \(0\). So:
\[
2xy + x^2y' + 3y^2y' = 0
\]
Step B — Factor and isolate \(y'\)
\[
2xy + y'(x^2 + 3y^2)=0
\]
\[
y'(x^2 + 3y^2) = -2xy
\]
\[
\boxed{\frac{dy}{dx} = -\frac{2xy}{x^2 + 3y^2}}
\]
3) Second derivative \(y''\)
To find \(y''\), differentiate the already differentiated equation again.
Since \(y'\) depends on \(x\), its derivative is:
\[
\frac{d}{dx}(y') = y''.
\]
After differentiating again, you collect \(y''\) terms and isolate:
\[
A(x,y,y') + B(x,y,y')\,y'' = 0
\quad \Rightarrow \quad
y'' = -\frac{A}{B}.
\]
The calculator additionally substitutes \(y'\) from the first-derivative result so the final \(y''\) is expressed only in \(x\) and \(y\).
4) Point evaluation and tangent line
Once you have \(\dfrac{dy}{dx}\), you can evaluate it at a specific point \((x_0,y_0)\).
The tangent line at that point is:
\[
y - y_0 = \left.\frac{dy}{dx}\right|_{(x_0,y_0)}(x - x_0).
\]
If the slope denominator is \(0\), the tangent is vertical (slope undefined) and the line is \(x=x_0\).
5) Graphing the implicit curve
The calculator plots the curve by rewriting the equation as:
\[
F(x,y)=0
\]
and drawing the contour where \(F\) changes sign (a standard contouring method called marching squares).
- Click on the graph to set \((x_0,y_0)\).
- The tangent line is drawn using the computed slope.
- Drag to pan; use wheel/pinch to zoom (large zoom-out is supported).
6) Input notes
- If you enter a single expression (without “=”), it is treated as “= 0”.
- Implicit multiplication is allowed:
x y, 2x, (x+1)(y-1).
- Shorthand like
cos^2(2x) is interpreted as \((\cos(2x))^2\).
7) Limitations
-
The solver isolates \(y'\) and \(y''\) assuming they appear linearly.
If a derivative ends up inside a denominator or exponent, the tool will report “nonlinear / cannot isolate”.
-
\(|u|\) is not differentiable where \(u=0\). The tool uses \(\frac{u}{|u|}u'\) away from those points.
