IB Mathematics AA HL Flashcards -Implicit differentiation

Up until now, we have only seen functions that have \( y \) on one side and all \( x \)’s on the other side [defines \( y \) explicitly as a function of \( x \)]. However, if we consider something such as \( x^3 + y^3 – 9xy = 0 \), it does define \( y \) in relation to \( x \), but only implicitly. If we assume this equation defines \( y \) as a differentiable function of \( x \), then \(\frac{dy}{dx}\) can still be found with a technique called implicit differentiation. Just remember \( y \) is a function of \( x \).

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METHOD:
1. Differentiate term-by-term, both sides, with respect to \( x \). Use the chain rule for any terms containing \( y \).
2. Collect all \( \frac{dy}{dx} \) terms on one side, all others on the other side.
3. Factor out \( \frac{dy}{dx} \).
4. Divide both sides by the factor.
5. Simplify as much as possible.

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E.G.: Differentiate \( x^3 + y^3 – 9xy = 0 \) implicitly:
1. TREATING THE \( y \)’s AS FUNCTIONS OF \( x \):
\(\frac{d}{dx}(x^3) + \frac{d}{dx}(y^3) – \frac{d}{dx}(9xy) = \frac{d}{dx}(0)\)
2. PRODUCT RULE FOR \( 9xy \), CHAIN RULE FOR \( y^3 \):
\(3x^2 + 3y^2\frac{dy}{dx} – 9\left(x\frac{dy}{dx} + y\frac{dx}{dx}\right) = 0\)
3. SIMPLIFY:
\(3x^2 + 3y^2\frac{dy}{dx} – 9x\frac{dy}{dx} – 9y = 0\)
4. Collect terms:
\(3y^2\frac{dy}{dx} – 9x\frac{dy}{dx} = 9y – 3x^2\)
5. Factor out \( \frac{dy}{dx} \):
\(\frac{dy}{dx}(3y^2 – 9x) = 9y – 3x^2\)
6. Solve for \( \frac{dy}{dx} \):
\(\frac{dy}{dx} = \frac{9y – 3x^2}{3y^2 – 9x}\)

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RELATED RATES
We might then see some situations where three variables are all changing at different rates, related to each of the other variables. This is essentially just another use of the chain rule \(\frac{dy}{dx} = \frac{dy}{db} \times \frac{db}{dx}\), almost like how \(\frac{da}{db}\). This requires info about relationships between 3 variables. You may also need to remember: If \(\frac{da}{db}\), then \(\frac{db}{da} = \frac{1}{a/b}\).

E.G. : A circle is growing at a rate of 20 cm/s. Find the rate of change of the area at the point when \( r = 50 \) cm:
1. The rate of change of the area is given by \(\frac{dA}{dt}\). However, we only know \( A \) in terms of \( r \) (\( A = \pi r^2 \)), so we use \(\frac{dA}{dt} = \frac{dA}{dr} \cdot \frac{dr}{dt}\) and differentiate \( A \) to find \(\frac{dA}{dr} = 2\pi r\). We were also told that \(\frac{dr}{dt} = 20\). Therefore, \(\frac{dA}{dt} = 2\pi r \cdot \frac{dr}{dt} = 2\pi \cdot 20\). When \( r = 50 \), \(\frac{dA}{dt} = 40\pi \cdot 50 = 2000\pi \approx 6280 \, \text{cm}^2/s\).

NOTE: Be prepared to use implicit differentiation within related rates questions. You may need to refer to some formulas in 3D geometry for these problems.

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