Lesson notes

This lesson solves two implicit-differentiation problems. The first proves an identity about reciprocal derivatives for a general conic, and the second differentiates a function built from two nested square roots.

Problem 1: an identity for the general conic

We are given the implicit equation

$ax^2 + 2hxy + by^2 = c^2$

and we want to prove that

$\frac{dy}{dx}\cdot\frac{dx}{dy} = 1.$

Differentiate with respect to $x$

Differentiating each term, and using the product rule on $2hxy$, gives

$2ax + 2h\left(x\frac{dy}{dx} + y\right) + 2by\frac{dy}{dx} = 0.$

The right-hand side is $0$ because $c^2$ is constant.

Solve for $\tfrac{dy}{dx}$

Collect the $\frac{dy}{dx}$ terms on one side and the rest on the other:

$\frac{dy}{dx}\,(2hx + 2by) = -2ax - 2hy.$

Dividing out the common factor of $2$,

$\frac{dy}{dx} = -\frac{ax + hy}{hx + by}.$

Form the product

Since $\frac{dx}{dy}$ is the reciprocal of $\frac{dy}{dx}$,

$\frac{dx}{dy} = -\frac{hx + by}{ax + hy}.$

Multiplying the two,

$\frac{dy}{dx}\cdot\frac{dx}{dy} = \left(-\frac{ax+hy}{hx+by}\right)\left(-\frac{hx+by}{ax+hy}\right) = 1.$

The two negatives give a positive, and the factors cancel, leaving $1$ as required.

Problem 2: a nested square-root function

We are given

$y = \sqrt{1 + \sqrt{1 + x^4}}$

and want to prove that

$y\,(y^2 - 1)\,\frac{dy}{dx} = x^3.$

Clear the radicals by squaring

Squaring once,

$y^2 = 1 + \sqrt{1 + x^4}, \qquad y^2 - 1 = \sqrt{1 + x^4}.$

Squaring again removes the remaining root:

$(y^2 - 1)^2 = 1 + x^4.$

Differentiate with respect to $x$

Using the chain rule on the left side and differentiating the right,

$2(y^2 - 1)\cdot 2y\,\frac{dy}{dx} = 4x^3,$

so

$4y\,(y^2 - 1)\,\frac{dy}{dx} = 4x^3.$

Dividing both sides by $4$,

$y\,(y^2 - 1)\,\frac{dy}{dx} = x^3,$

which is exactly what we set out to prove.

Key takeaways

• Differentiating $ax^2 + 2hxy + by^2 = c^2$ implicitly gives $\frac{dy}{dx} = -\dfrac{ax + hy}{hx + by}$.
• Because $\frac{dx}{dy}$ is the reciprocal of $\frac{dy}{dx}$, their product is always $1$.
• Squaring repeatedly to clear nested radicals, then differentiating with the chain rule, turns a messy root into a clean relation.