1.2 Differentiation Rules, Implicit & Inverse Differentiation

1.2 Differentiation Rules, Implicit & Inverse Differentiation

Differentiation measures local rates of change. The GRE Mathematics Subject Test evaluates both mechanical proficiency with differentiation formulas and theoretical mastery of differentiability conditions and higher-order derivative properties.

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Difference Quotients and Differentiability

Let $f$ be defined on an open interval around $a$. The derivative of $f$ at $a$ is: $f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h} = \lim_{x \to a} \frac{f(x) - f(a)}{x - a},$ provided this limit exists in $\mathbb{R}$.

Differentiability Implies Continuity

If $f$ is differentiable at $a$, it is continuous at $a$: $\lim_{x \to a} [f(x) - f(a)] = \lim_{x \to a} \left[ \frac{f(x) - f(a)}{x - a} (x - a) \right] = f'(a) \cdot 0 = 0.$ Hence $\lim_{x \to a} f(x) = f(a)$.

Canonical Non-Differentiable Pathologies

1. Corners / Sharp Turns: $f(x) = |x|$ at $x = 0$ has left derivative $-1$ and right derivative $+1$.
2. Vertical Tangents: $f(x) = x^{1/3}$ at $x = 0$ has difference quotient $h^{-2/3} \to \infty$.
3. Cusps: $f(x) = x^{2/3}$ at $x = 0$ has left quotient $\to -\infty$ and right quotient $\to +\infty$.
4. Differentiable with Discontinuous Derivative: $f(x) = \begin{cases} x^2 \sin(1/x) & x \neq 0 \\ 0 & x = 0 \end{cases}$
   • $f'(0) = \lim_{h \to 0} \frac{h^2 \sin(1/h)}{h} = 0$ by the Squeeze Theorem.
   • For $x \neq 0$, $f'(x) = 2x \sin(1/x) - \cos(1/x)$.
   • As $x \to 0$, $\cos(1/x)$ oscillates indefinitely. Therefore $\lim_{x \to 0} f'(x)$ does not exist, proving $f'$ is discontinuous at $0$ despite $f$ being differentiable everywhere.

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Differentiation Rules & Logarithmic Differentiation

Standard rules: $(fg)' = f'g + fg'$, $(f/g)' = (f'g - fg')/g^2$, and chain rule $(f \circ g)'(x) = f'(g(x)) g'(x)$.

Logarithmic Differentiation

For variable-base variable-exponent expressions $y = u(x)^{v(x)}$ ($u(x) > 0$), take logarithms before differentiating: $\ln y = v(x) \ln u(x) \implies \frac{y'}{y} = v'(x) \ln u(x) + \frac{v(x) u'(x)}{u(x)}.$ Solving for $y'$ gives $y' = u(x)^{v(x)} [v' \ln u + v u' / u]$. This also speeds up multivariable quotients: $\ln(f_1^{a_1} / g_1^{b_1}) \implies y'/y = a_1 f_1'/f_1 - b_1 g_1'/g_1$.

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Inverse Functions and Implicit Differentiation

Inverse Function Derivatives

If $f$ is strictly monotonic and differentiable with $f'(x) \neq 0$, then $f^{-1}$ is differentiable on its range. For $b = f(a)$ (so $a = f^{-1}(b)$): $(f^{-1})'(b) = \frac{1}{f'(f^{-1}(b))} = \frac{1}{f'(a)}.$

Standard Inverse Trigonometric Derivatives

1. $\frac{d}{dx} \arcsin x = \frac{1}{\sqrt{1 - x^2}}, \quad |x| < 1$
2. $\frac{d}{dx} \arccos x = -\frac{1}{\sqrt{1 - x^2}}, \quad |x| < 1$
3. $\frac{d}{dx} \arctan x = \frac{1}{1 + x^2}, \quad x \in \mathbb{R}$

Implicit Differentiation on Curves

For level curves $F(x, y) = 0$, applying the chain rule gives: $F_x + F_y \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{F_x}{F_y} \quad (F_y \neq 0).$

• Horizontal Tangents: $F_x = 0$ with $F_y \neq 0$.
• Vertical Tangents: $F_y = 0$ with $F_x \neq 0$.
• Singular Points: $F_x = 0$ and $F_y = 0$ simultaneously.

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Leibniz's Rule for Higher-Order Derivatives

For $n$-times differentiable functions $u$ and $v$, the product rule generalizes as: $(u \cdot v)^{(n)} = \sum_{k=0}^n \binom{n}{k} u^{(n-k)} v^{(k)}.$ When $v(x)$ is a polynomial of degree $m$, all derivatives $v^{(k)}$ vanish for $k > m$, terminating the sum after $m+1$ terms.

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Step-by-Step Worked Problems

Problem 1: Higher Derivatives via Leibniz

Compute $f^{(15)}(0)$ for $f(x) = (x^2 - 2x) e^{-x}$.

Solution:

1. Let $u(x) = e^{-x}$ and $v(x) = x^2 - 2x$.
2. Derivatives of $u$: $u^{(m)}(0) = (-1)^m$.
3. Derivatives of $v$: $v(0) = 0$, $v'(0) = -2$, $v''(0) = 2$, and $v^{(k)}(0) = 0$ for $k \ge 3$.
4. Applying Leibniz's rule with $n = 15$: $f^{(15)}(0) = \binom{15}{0} u^{(15)}(0) v(0) + \binom{15}{1} u^{(14)}(0) v'(0) + \binom{15}{2} u^{(13)}(0) v''(0).$
5. Substitute evaluations: $f^{(15)}(0) = 1(-1)(0) + 15(1)(-2) + \frac{15 \times 14}{2}(-1)(2) = 0 - 30 - 210 = -240.$

Problem 2: Implicit Differentiation for Tangents

Find the tangent line to $x^3 + y^3 = 2xy$ at $(1, 1)$.

Solution:

1. Differentiate implicitly: $3x^2 + 3y^2 y' = 2y + 2x y'$.
2. Group terms: $y'(3y^2 - 2x) = 2y - 3x^2 \implies y' = \frac{2y - 3x^2}{3y^2 - 2x}$.
3. Substitute $(1, 1)$: $y' = \frac{2(1) - 3(1)}{3(1) - 2(1)} = \frac{-1}{1} = -1$.
4. The tangent line equation is $y - 1 = -1(x - 1) \implies y = -x + 2$.

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GRE Exam Traps & Pitfalls

Trap 1: Evaluating Inverse Derivatives at Target Rather than Preimage When computing $(f^{-1})'(b)$, do not evaluate $1/f'(b)$. First solve $f(a) = b$ for $a$, then calculate $1/f'(a)$.

Trap 2: Dropping Product Rule in Mixed Implicit Terms When taking $\frac{d}{dx}$ of $xy$, always write $y + x y'$. Forgetting the $y$ term is a ubiquitous calculation error under exam pressure.

Trap 3: Assuming $f'$ Is Continuous Because $f$ Is Differentiable Derivatives satisfy the Intermediate Value Property (Darboux's Theorem) but can have essential oscillatory discontinuities.