Implicit Differentiation – Quick Review

Key Idea: Implicit Differentiation

• Used when $y$ is not isolated (relation, not explicit function).
• Treat $y$ as an unknown function of $x$. Whenever you differentiate a term containing $y$, multiply by $\frac{dy}{dx}$ (chain rule).
• Apply familiar rules (power, product, quotient, chain) exactly as for explicit functions.
• After differentiating both sides, algebraically solve for $\frac{dy}{dx}$.

Core Procedure

• Differentiate both sides of the given equation with respect to $x$.
• Attach $\frac{dy}{dx}$ to every $y$–derivative.
• Use product/chain rule wherever $x$ and $y$ are multiplied or nested.
• Collect all $\frac{dy}{dx}$ terms on one side.
• Factor out $\frac{dy}{dx}$.
• Isolate by dividing: $\frac{dy}{dx}=\dfrac{\text{terms without }\frac{dy}{dx}}{\text{coefficient of }\frac{dy}{dx}}$.

Quick Reference: Product & Chain Rules

• Product rule: $(uv)' = u'v + uv'$.
• Chain rule with $y$: If $f(y)$, then $\dfrac{d}{dx}f(y)=f'(y)\frac{dy}{dx}$.

Worked-Out Results (for rapid recall)

• $x^2 = 3y^4 + x \;\Rightarrow\; \displaystyle \frac{dy}{dx}=\frac{2x-1}{12y^3}$
• $e^{2x}=x^3y^4 \;\Rightarrow\; \displaystyle \frac{dy}{dx}=\frac{2e^{2x}-3x^2y^4}{4x^3y^3}$
• $x^5+3x^2y^3+y^5=2 \;\Rightarrow\; \displaystyle \frac{dy}{dx}=\frac{-5x^4-6xy^3}{9x^2y^2+5y^4}$
• $3x^3y^4-\ln(5x^3y)=5 \;\Rightarrow\; \displaystyle \frac{dy}{dx}=\frac{15x^2y-45x^3y^5}{60x^3y^3+5x^2}$

Common Pitfalls

• Forgetting the extra $\frac{dy}{dx}$ factor when differentiating any $y$ term.
• Skipping the product rule when $x$ and $y$ appear in the same factor (e.g., $x^3y^4$).
• Failing to distribute negative signs or combine like terms before factoring $\frac{dy}{dx}$.

Rapid Checklist Before Finishing

• [ ] Did every $y$-derivative get a $\frac{dy}{dx}$?
• [ ] Were product/chain rules correctly applied?
• [ ] Are all $\frac{dy}{dx}$ terms on one side and factored?
• [ ] Final answer solved explicitly for $\frac{dy}{dx}$?