UNIT 3: DIFFERENTIATION: COMPOSITE, IMPLICIT, AND INVERSE FUNCTIONS

1. INTRODUCTION

Ready to level up? So far, we've handled basic functions, but real life is rarely that simple—it’s layered. Think of a ripple in a pond: the area depends on the radius, which depends on time. That's a "function within a function," and we need the Chain Rule to crack it.

In this unit, we’re moving past $y = f(x)$ to tackle messy equations where $x$ and $y$ are "it's complicated" (Implicit Differentiation) and flipping the script with Inverse Functions. By the end, you'll be able to differentiate the complex, multi-layered models used in high-level STEM and economics. Let's dive in.

2. ALL KEY CONCEPTS, TERMS, AND PRINCIPLES

Key Terms and Definitions

• Composite Function: A function $h(x) = f(g(x))$ where the output of the inner function $g(x)$ becomes the input for the outer function $f(x)$.
• The Chain Rule: The formula used to find the derivative of a composite function. It states that the derivative of $f(g(x))$ is $f'(g(x)) \cdot g'(x)$.
• Implicit Function: A function where the dependent variable $y$ is not isolated on one side of the equation (e.g., $\sin(xy) = x + y$).
• Explicit Function: A function where $y$ is expressed directly in terms of $x$ (e.g., $y = x^2 + 2x$).
• Inverse Function ($f^{-1}$): A function that "reverses" another function. If $f(a) = b$, then $f^{-1}(b) = a$.
• Inverse Trigonometric Functions: The inverses of the trigonometric ratios, such as $\arcsin(x)$, $\arccos(x)$, and $\arctan(x)$.
• Horizontal Tangent: A point on a curve where the derivative $\frac{dy}{dx} = 0$.
• Vertical Tangent: A point on a curve where the derivative $\frac{dy}{dx}$ is undefined (specifically, where the denominator of the derivative expression is zero while the numerator is non-zero).

Foundational Principles

1. Leibniz Notation for the Chain Rule: If $y = f(u)$ and $u = g(x)$, then: $$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$$
2. Implicit Differentiation Principle: When differentiating with respect to $x$, any term involving $y$ must be treated as a composition $y(x)$, necessitating the use of the Chain Rule (multiplying by $\frac{dy}{dx}$).
3. Derivative of an Inverse Function: If $g(x)$ is the inverse of $f(x)$, then: $$g'(x) = \frac{1}{f'(g(x))}$$ Or, at a specific point $(a, b)$ on $f$ and $(b, a)$ on $f^{-1}$: $$(f^{-1})'(b) = \frac{1}{f'(a)}$$

3. IN-DEPTH EXPLANATION

3.1 The Chain Rule (Level 10-30)

The Chain Rule is the most widely used differentiation rule in calculus. It allows us to differentiate "layered" functions. Think of it like a processing line:

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The Formula: If $y = f(u)$ and $u = g(x)$, then the derivative of $y$ with respect to $x$ is: $$\frac{dy}{dx} = f'(g(x)) \cdot g'(x)$$

Step-by-Step Application:

1. Identify the layers: Determine the "outer" function $f$ and the "inner" function $g$.
2. Differentiate the outer layer: Take the derivative of $f$ while keeping $g(x)$ inside it untouched.
3. Differentiate the inner layer: Take the derivative of $g(x)$.
4. Multiply: The final result is the product of the two derivatives.

Example: To differentiate $y = (3x^2 + 1)^5$:

• Outer: $u^5$, derivative is $5u^4$.
• Inner: $3x^2 + 1$, derivative is $6x$.
• Result: $5(3x^2 + 1)^4 \cdot (6x) = 30x(3x^2 + 1)^4$.

3.2 Implicit Differentiation (Level 40-60)

Most functions we have encountered are explicit, like $y = \sqrt{x}$. However, equations like $x^2 + y^2 = 25$ (a circle) are implicit. We could solve for $y = \pm\sqrt{25 - x^2}$, but this creates two separate functions. Implicit differentiation allows us to find the slope $\frac{dy}{dx}$ without isolating $y$.

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The Process:

1. Differentiate both sides of the equation with respect to $x$.
2. Apply the Power/Product/Quotient rules as usual for $x$.
3. Apply the Chain Rule for $y$: every time you differentiate a term containing $y$, multiply by $\frac{dy}{dx}$.
4. Collect all terms containing $\frac{dy}{dx}$ on one side.
5. Factor out $\frac{dy}{dx}$ and solve.

Example: $x^2 + y^2 = 25$ $$\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(25)$$ $$2x + 2y \frac{dy}{dx} = 0$$ $$2y \frac{dy}{dx} = -2x \implies \frac{dy}{dx} = -\frac{x}{y}$$

3.3 Derivatives of Inverse Functions (Level 60-80)

The derivative of an inverse function is geometrically related to the derivative of the original function. Since an inverse function is a reflection over the line $y = x$, the slope at a point $(a, b)$ on $f$ is the reciprocal of the slope at point $(b, a)$ on $f^{-1}$.

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The Theorem: If $f$ is differentiable and has an inverse $g$, then: $$g'(x) = \frac{1}{f'(g(x))}$$ This is derived by starting with $f(g(x)) = x$ and differentiating both sides using the chain rule: $$f'(g(x)) \cdot g'(x) = 1 \implies g'(x) = \frac{1}{f'(g(x))}$$

3.4 Derivatives of Inverse Trigonometric Functions (Level 80-90)

Using implicit differentiation, we can derive the derivatives of inverse trig functions. Let $y = \arcsin(x)$. This means $\sin(y) = x$. Differentiating implicitly: $$\cos(y) \cdot \frac{dy}{dx} = 1 \implies \frac{dy}{dx} = \frac{1}{\cos(y)}$$ Using the identity $\cos^2(y) + \sin^2(y) = 1$, we know $\cos(y) = \sqrt{1 - \sin^2(y)} = \sqrt{1 - x^2}$. Thus: $$\frac{d}{dx}(\arcsin x) = \frac{1}{\sqrt{1 - x^2}}$$

Standard Derivatives Table:

3.5 Higher-Order Derivatives of Implicit Functions (Level 90-100)

To find the second derivative $\frac{d^2y}{dx^2}$ of an implicitly defined curve, differentiate the first derivative expression $\frac{dy}{dx}$ with respect to $x$. This usually involves the quotient rule and requires substituting the expression for $\frac{dy}{dx}$ back into the final equation to get the result in terms of $x$ and $y$.

4. EXAMPLES

Example 1: Basic Chain Rule

Differentiate $y = \sin(x^3)$. Solution: Inner function $u = x^3$, $u' = 3x^2$. Outer function $f(u) = \sin(u)$, $f'(u) = \cos(u)$. $$\frac{dy}{dx} = \cos(x^3) \cdot 3x^2 = 3x^2 \cos(x^3)$$

Example 2: Power Rule with Chain Rule

Differentiate $f(x) = (4x^2 - 3x + 2)^7$. Solution: $$f'(x) = 7(4x^2 - 3x + 2)^6 \cdot \frac{d}{dx}(4x^2 - 3x + 2)$$ $$f'(x) = 7(4x^2 - 3x + 2)^6 (8x - 3)$$

Example 3: Chain Rule with Exponential Functions

Differentiate $y = e^{\tan x}$. Solution: The derivative of $e^u$ is $e^u \cdot \frac{du}{dx}$. $$\frac{dy}{dx} = e^{\tan x} \cdot \sec^2 x$$

Example 4: Nested Chain Rule (Level 50)

Differentiate $y = \ln(\cos(e^x))$. Solution: This has three layers: $\ln(\dots)$, $\cos(\dots)$, and $e^x$. $$\frac{dy}{dx} = \frac{1}{\cos(e^x)} \cdot (-\sin(e^x)) \cdot e^x$$ $$\frac{dy}{dx} = -e^x \tan(e^x)$$

Example 5: Basic Implicit Differentiation

Find $\frac{dy}{dx}$ for $x^3 + y^3 = 6xy$. Solution: Differentiate with respect to $x$: $$3x^2 + 3y^2 \frac{dy}{dx} = 6(1 \cdot y + x \cdot \frac{dy}{dx})$$ $$3x^2 + 3y^2 \frac{dy}{dx} = 6y + 6x \frac{dy}{dx}$$ $$(3y^2 - 6x) \frac{dy}{dx} = 6y - 3x^2$$ $$\frac{dy}{dx} = \frac{6y - 3x^2}{3y^2 - 6x} = \frac{2y - x^2}{y^2 - 2x}$$

Example 6: Implicit Differentiation with Trig

Find $\frac{dy}{dx}$ for $\sin(x+y) = y$. Solution: $$\cos(x+y) \cdot (1 + \frac{dy}{dx}) = \frac{dy}{dx}$$ $$\cos(x+y) + \cos(x+y)\frac{dy}{dx} = \frac{dy}{dx}$$ $$\cos(x+y) = \frac{dy}{dx} - \cos(x+y)\frac{dy}{dx}$$ $$\frac{dy}{dx} = \frac{\cos(x+y)}{1 - \cos(x+y)}$$

Example 7: Tangent Line to an Implicit Curve

Find the equation of the tangent line to $x^2 + xy + y^2 = 7$ at the point $(1, 2)$. Solution:

1. Differentiate: $2x + (y + x\frac{dy}{dx}) + 2y\frac{dy}{dx} = 0$.
2. Plug in $(1, 2)$: $2(1) + 2 + 1\frac{dy}{dx} + 2(2)\frac{dy}{dx} = 0$.
3. $4 + 5\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -4/5$.
4. Line equation: $y - 2 = -\frac{4}{5}(x - 1)$.

Example 8: Horizontal and Vertical Tangents

For the ellipse $4x^2 + 9y^2 = 36$, find the points where the tangent is horizontal. Solution: Differentiate: $8x + 18y \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{4x}{9y}$. For horizontal tangent, $\frac{dy}{dx} = 0 \implies -4x = 0 \implies x = 0$. Plug $x = 0$ into original: $9y^2 = 36 \implies y^2 = 4 \implies y = \pm 2$. Points: $(0, 2)$ and $(0, -2)$.

Example 9: Vertical Tangents

For the same ellipse $4x^2 + 9y^2 = 36$, find points where the tangent is vertical. Solution: Tangent is vertical where the denominator of $\frac{dy}{dx}$ is $0$: $9y = 0 \implies y = 0$. Plug $y = 0$ into original: $4x^2 = 36 \implies x^2 = 9 \implies x = \pm 3$. Points: $(3, 0)$ and $(-3, 0)$.

Example 10: Inverse Function Derivative (Numerical)

If $f(x) = x^3 + x$, find $(f^{-1})'(2)$. Solution:

1. Let $g = f^{-1}$. We need $g'(2)$.
2. Find $x$ such that $f(x) = 2$: $x^3 + x = 2 \implies x = 1$ (by inspection).
3. $f'(x) = 3x^2 + 1$.
4. $g'(2) = \frac{1}{f'(1)} = \frac{1}{3(1)^2 + 1} = \frac{1}{4}$.

Example 11: Inverse Function Derivative (Algebraic)

Find the derivative of the inverse of $f(x) = \sqrt{x-1}$. Solution: $y = \sqrt{x-1} \implies x = y^2 + 1$. So $f^{-1}(x) = x^2 + 1$. $\frac{d}{dx}(f^{-1}(x)) = 2x$. Using the formula: $f'(x) = \frac{1}{2\sqrt{x-1}}$. $g'(x) = \frac{1}{f'(g(x))} = \frac{1}{\frac{1}{2\sqrt{(x^2+1)-1}}} = 2\sqrt{x^2} = 2x$. (Matches!)

Example 12: Derivative of $\arctan(x^2)$

Find $f'(x)$ for $f(x) = \arctan(x^2)$. Solution: Using Chain Rule on $\arctan(u)$: $$f'(x) = \frac{1}{1 + (x^2)^2} \cdot \frac{d}{dx}(x^2) = \frac{2x}{1 + x^4}$$

Example 13: Derivative of $\arcsin(e^x)$

Find $\frac{dy}{dx}$ for $y = \arcsin(e^x)$. Solution: $$\frac{dy}{dx} = \frac{1}{\sqrt{1 - (e^x)^2}} \cdot e^x = \frac{e^x}{\sqrt{1 - e^{2x}}}$$

Example 14: Implicit Differentiation with Second Derivative (Level 95)

Find $\frac{d^2y}{dx^2}$ for $x^2 + y^2 = 1$. Solution:

1. $\frac{dy}{dx} = -x/y$.
2. $\frac{d^2y}{dx^2} = \frac{d}{dx}(-x/y) = \frac{(-1)(y) - (-x)\frac{dy}{dx}}{y^2}$ (Quotient Rule).
3. Substitute $\frac{dy}{dx} = -x/y$: $$\frac{d^2y}{dx^2} = \frac{-y + x(-x/y)}{y^2} = \frac{-y - x^2/y}{y^2} = \frac{-y^2 - x^2}{y^3}$$
4. Since $x^2 + y^2 = 1$, then $-y^2 - x^2 = -1$. $$\frac{d^2y}{dx^2} = -\frac{1}{y^3}$$

Example 15: Logarithmic and Chain Rule Combined

Differentiate $y = \ln(x^2 + 2x + 1)$. Solution: $$y' = \frac{1}{x^2 + 2x + 1} \cdot (2x + 2) = \frac{2(x+1)}{(x+1)^2} = \frac{2}{x+1}$$

Example 16: Derivative of $\text{arcsec}(5x)$

Find $f'(x)$ for $f(x) = \text{arcsec}(5x)$. Solution: $$f'(x) = \frac{1}{|5x|\sqrt{(5x)^2 - 1}} \cdot 5 = \frac{5}{5|x|\sqrt{25x^2 - 1}} = \frac{1}{|x|\sqrt{25x^2 - 1}}$$

Example 17: Product and Chain Rule

Differentiate $y = x^2 \cos(3x)$. Solution: $$y' = (x^2)' \cos(3x) + x^2 (\cos(3x))'$$ $$y' = 2x \cos(3x) + x^2 (-\sin(3x) \cdot 3) = 2x \cos(3x) - 3x^2 \sin(3x)$$

Example 18: Implicit Differentiation with $e^y$

Find $\frac{dy}{dx}$ if $e^y = x^2 + y$. Solution: $$e^y \frac{dy}{dx} = 2x + \frac{dy}{dx}$$ $$(e^y - 1) \frac{dy}{dx} = 2x \implies \frac{dy}{dx} = \frac{2x}{e^y - 1}$$

Example 19: Inverse Trig with Quotient Rule

Differentiate $y = \frac{\arctan x}{x}$. Solution: $$y' = \frac{(\frac{1}{1+x^2})(x) - (1)(\arctan x)}{x^2} = \frac{x - (1+x^2)\arctan x}{x^2(1+x^2)}$$

Example 20: Finding slope at a point (Implicit)

Find the slope of $y^2(y^2-4) = x^2(x^2-5)$ at $(0, -2)$. Solution: $$(4y^3 - 8y)\frac{dy}{dx} = 4x^3 - 10x$$ Plug in $x=0, y=-2$: $$(4(-8) - 8(-2))\frac{dy}{dx} = 0$$ $$(-32 + 16)\frac{dy}{dx} = 0 \implies -16\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = 0$$

Example 21: Complex Chain Rule with Square Roots

Differentiate $y = \sqrt{1 + \sqrt{x}}$. Solution: $$y' = \frac{1}{2\sqrt{1 + \sqrt{x}}} \cdot \frac{d}{dx}(1 + \sqrt{x})$$ $$y' = \frac{1}{2\sqrt{1 + \sqrt{x}}} \cdot \frac{1}{2\sqrt{x}} = \frac{1}{4\sqrt{x + x\sqrt{x}}}$$

Example 22: Derivative of $\arccos(\sin x)$

Find $\frac{dy}{dx}$ for $y = \arccos(\sin x)$. Solution: $$\frac{dy}{dx} = \frac{-1}{\sqrt{1 - \sin^2 x}} \cdot \cos x = \frac{-1}{\sqrt{\cos^2 x}} \cdot \cos x = \frac{-\cos x}{|\cos x|}$$ (Result is $-1$ or $1$ depending on the quadrant).

Example 23: Implicit with Product Rule and Power

Find $\frac{dy}{dx}$ for $(x+y)^3 = x^3 + y^3$. Solution: $$3(x+y)^2(1 + \frac{dy}{dx}) = 3x^2 + 3y^2\frac{dy}{dx}$$ $$(x+y)^2 + (x+y)^2\frac{dy}{dx} = x^2 + y^2\frac{dy}{dx}$$ $$\frac{dy}{dx}((x+y)^2 - y^2) = x^2 - (x+y)^2$$ $$\frac{dy}{dx} = \frac{x^2 - (x^2+2xy+y^2)}{(x^2+2xy+y^2) - y^2} = \frac{-2xy-y^2}{x^2+2xy}$$

Example 24: Inverse Function at a Specific Point (Table Based)

Given the table:

Find $(f^{-1})'(3)$. Solution: We need $x$ such that $f(x) = 3$. From the table, $x=1$. $$(f^{-1})'(3) = \frac{1}{f'(1)} = \frac{1}{5}$$

Example 25: Triple Composition Chain Rule

Differentiate $y = \cos^4(x^2)$. Solution: Layers: $u^4 \rightarrow \cos(v) \rightarrow x^2$. $$y' = 4\cos^3(x^2) \cdot (-\sin(x^2)) \cdot 2x = -8x \cos^3(x^2) \sin(x^2)$$

Example 26: Derivative of Inverse Trig with Chain Rule

Differentiate $y = \text{arccot}(\sqrt{x})$. Solution: $$y' = \frac{-1}{1 + (\sqrt{x})^2} \cdot \frac{1}{2\sqrt{x}} = \frac{-1}{2\sqrt{x}(1+x)}$$

Example 27: Evaluating Second Derivative Implicitly

Find $\frac{d^2y}{dx^2}$ at $(1,1)$ for $x^2 + y^2 = 2x$. Solution:

1. $2x + 2y y' = 2 \implies y' = \frac{2 - 2x}{2y} = \frac{1-x}{y}$.
2. At $(1,1)$, $y' = \frac{1-1}{1} = 0$.
3. $y'' = \frac{(-1)(y) - (1-x)y'}{y^2}$.
4. At $(1,1)$ with $y'=0$: $y'' = \frac{-1 - (0)(0)}{1^2} = -1$.

Example 28: Chain Rule in a Physics Context

The volume of a sphere is $V = \frac{4}{3}\pi r^3$. If the radius is changing with time $r(t)$, find $\frac{dV}{dt}$. Solution: By the Chain Rule: $$\frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dt} = 4\pi r^2 \frac{dr}{dt}$$

Example 29: Derivative of $y = \ln(\sec x + \tan x)$

Solution: $$y' = \frac{1}{\sec x + \tan x} \cdot (\sec x \tan x + \sec^2 x)$$ $$y' = \frac{\sec x (\tan x + \sec x)}{\sec x + \tan x} = \sec x$$

Example 30: Derivative of $x = \sin y$ (Inverse Trig Derivation style)

Find $\frac{dy}{dx}$. Solution: $$1 = \cos y \frac{dy}{dx} \implies \frac{dy}{dx} = \sec y = \frac{1}{\sqrt{1 - \sin^2 y}} = \frac{1}{\sqrt{1 - x^2}}$$

Example 31: Advanced Implicit Product Rule

Find $\frac{dy}{dx}$ for $x^2y^2 = (x+y)^2$. Solution: $2xy^2 + x^2(2y y') = 2(x+y)(1+y')$. $2xy^2 + 2x^2y y' = 2x + 2y + 2x y' + 2y y'$. $x^2y y' - x y' - y y' = x + y - xy^2$. $y'(x^2y - x - y) = x + y - xy^2$. $y' = \frac{x + y - xy^2}{x^2y - x - y}$.

Example 32: Derivative of $y = \arcsin(x/a)$

Solution: $$y' = \frac{1}{\sqrt{1 - (x/a)^2}} \cdot \frac{1}{a} = \frac{1}{\sqrt{\frac{a^2 - x^2}{a^2}}} \cdot \frac{1}{a} = \frac{1}{\frac{\sqrt{a^2 - x^2}}{a}} \cdot \frac{1}{a} = \frac{1}{\sqrt{a^2 - x^2}}$$

Example 33: Chain Rule with Absolute Value

Find the derivative of $y = \ln|x|$. Solution: For $x > 0$, $y = \ln x$, $y' = 1/x$. For $x < 0$, $y = \ln(-x)$, $y' = \frac{1}{-x} \cdot (-1) = 1/x$. Thus, for all $x \neq 0$, $\frac{d}{dx} \ln|x| = \frac{1}{x}$.

Example 34: Inverse Function of a Linear-Rational Function

Find $(f^{-1})'(y)$ if $f(x) = \frac{x-1}{x+1}$. Solution: $f'(x) = \frac{(1)(x+1) - (1)(x-1)}{(x+1)^2} = \frac{2}{(x+1)^2}$. $(f^{-1})'(y) = \frac{1}{f'(x)} = \frac{(x+1)^2}{2}$. To express in terms of $y$, solve $y = \frac{x-1}{x+1} \implies yx + y = x - 1 \implies x(y-1) = -y-1 \implies x = \frac{y+1}{1-y}$. Substitute $x$: $(f^{-1})'(y) = \frac{(\frac{y+1}{1-y} + 1)^2}{2} = \dots = \frac{2}{(1-y)^2}$.

Example 35: High Level Implicit Second Derivative

Show that for the circle $x^2 + y^2 = r^2$, the curvature-related value $y''$ satisfies $y'' = -\frac{r^2}{y^3}$. Solution: As shown in Example 14, $y' = -x/y$ and $y'' = \frac{-y^2 - x^2}{y^3}$. Substituting $x^2 + y^2 = r^2$: $$y'' = \frac{-(x^2+y^2)}{y^3} = -\frac{r^2}{y^3}$$

End of Chapter.