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Home/Calculus/Chapter 4/Chain Rule

CALC-4.2

4-5 hours

Chain Rule and Higher-Order Derivatives

Master the chain rule for composite functions, learn to differentiate inverse functions, implicit equations, parametric curves, and explore higher-order derivatives.

Learning Objectives

Apply the chain rule to differentiate composite functions

Derive and use the formula for derivatives of inverse functions

Perform implicit differentiation on equations defining y implicitly

Compute derivatives of functions given by parametric equations

Calculate higher-order derivatives of various functions

Apply Leibniz's formula for nth derivatives of products

Understand the relationship between higher differentials and higher derivatives

Solve complex differentiation problems combining multiple techniques

1. The Chain Rule

The chain rule is one of the most important rules in calculus. It tells us how to differentiate composite functions—functions built by putting one function inside another.

Theorem 4.4: Chain Rule

Let $y = f(u)$ be differentiable at $u_0$ , and $u = g(x)$ be differentiable at $x_0$ with $u_0 = g(x_0)$ . Then the composite function $F(x) = f(g(x))$ is differentiable at $x_0$ , and:

F'(x_0) = f'(g(x_0)) \cdot g'(x_0)

In Leibniz notation: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

Proof of Theorem 4.4:

We compute the difference quotient:

\frac{F(x) - F(x_0)}{x - x_0} = \frac{f(g(x)) - f(g(x_0))}{x - x_0}

When $g(x) \neq g(x_0)$ , we can write:

= \frac{f(g(x)) - f(g(x_0))}{g(x) - g(x_0)} \cdot \frac{g(x) - g(x_0)}{x - x_0}

As $x \to x_0$ : the first factor → $f'(g(x_0))$ , the second → $g'(x_0)$ .

∎

Example 4.5: Chain Rule Application

Find the derivative of $y = \sin(x^2)$ .

Let $u = x^2$ , so $y = \sin u$ .

\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \cos u \cdot 2x = 2x\cos(x^2)

Example 4.6: Nested Chain Rule

Find $\frac{d}{dx}e^{\sin(x^2)}$ .

Apply chain rule twice:

\frac{d}{dx}e^{\sin(x^2)} = e^{\sin(x^2)} \cdot \cos(x^2) \cdot 2x = 2x\cos(x^2)e^{\sin(x^2)}

Remark 4.1: Chain Rule in Different Notations

The chain rule can be expressed in several equivalent ways:

Leibniz Notation

\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}

Prime Notation

(f \circ g)'(x) = f'(g(x)) \cdot g'(x)

Operator Notation

D_x[f(g(x))] = D_u f(u)|_{u=g(x)} \cdot D_x g(x)

Differential Notation

df = f'(u)\, du = f'(g(x)) \cdot g'(x)\, dx

Example 4.6b: Chain Rule with Trigonometric Functions

Problem: Differentiate $y = \tan^3(2x)$ .

Solution:

Let $u = \tan(2x)$ , so $y = u^3$ .

\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 3u^2 \cdot \sec^2(2x) \cdot 2

= 6\tan^2(2x)\sec^2(2x)

Example 4.6c: Chain Rule with Logarithms

Problem: Differentiate $y = \ln(\sin x)$ for $x \in (0, \pi)$ .

Solution:

\frac{dy}{dx} = \frac{1}{\sin x} \cdot \cos x = \cot x

Note: We need $\sin x > 0$ , which holds for $x \in (0, \pi)$ .

Example 4.6d: Chain Rule with Square Roots

Problem: Differentiate $y = \sqrt{1 + x^2}$ .

Solution:

Write as $y = (1 + x^2)^{1/2}$ .

\frac{dy}{dx} = \frac{1}{2}(1 + x^2)^{-1/2} \cdot 2x = \frac{x}{\sqrt{1 + x^2}}

Theorem 4.4b: Generalized Chain Rule

For a composition of three or more functions $y = f(g(h(x)))$ :

\frac{dy}{dx} = f'(g(h(x))) \cdot g'(h(x)) \cdot h'(x)

In Leibniz notation with $u = h(x)$ , $v = g(u)$ , $y = f(v)$ :

\frac{dy}{dx} = \frac{dy}{dv} \cdot \frac{dv}{du} \cdot \frac{du}{dx}

Example 4.6e: Triple Chain Rule

Problem: Find $\frac{d}{dx}\cos(e^{\sqrt{x}})$ .

Solution:

Let $u = \sqrt{x}$ , $v = e^u$ , $y = \cos v$ .

\frac{dy}{dx} = -\sin(e^{\sqrt{x}}) \cdot e^{\sqrt{x}} \cdot \frac{1}{2\sqrt{x}}

= -\frac{e^{\sqrt{x}}\sin(e^{\sqrt{x}})}{2\sqrt{x}}

Remark 4.1b: Common Chain Rule Patterns

Memorize these common patterns to speed up differentiation:

Function	Derivative
$[f(x)]^n$	$n[f(x)]^{n-1} \cdot f'(x)$
$e^{f(x)}$	$e^{f(x)} \cdot f'(x)$
$\ln f(x)$	$\frac{f'(x)}{f(x)}$
$\sin f(x)$	$\cos f(x) \cdot f'(x)$
$\cos f(x)$	$-\sin f(x) \cdot f'(x)$
$\tan f(x)$	$\sec^2 f(x) \cdot f'(x)$
$a^{f(x)}$	$a^{f(x)} \ln a \cdot f'(x)$

Example 4.6f: Combining Chain Rule with Other Rules

Problem: Differentiate $y = \frac{e^{x^2}}{\sin x}$ .

Solution:

Use quotient rule combined with chain rule:

\frac{dy}{dx} = \frac{(e^{x^2})' \cdot \sin x - e^{x^2} \cdot (\sin x)'}{\sin^2 x}

The derivative of $e^{x^2}$ requires chain rule:

(e^{x^2})' = e^{x^2} \cdot 2x

Substituting:

\frac{dy}{dx} = \frac{2xe^{x^2}\sin x - e^{x^2}\cos x}{\sin^2 x} = \frac{e^{x^2}(2x\sin x - \cos x)}{\sin^2 x}

Example 4.6g: Exponential of Exponential

Problem: Differentiate $y = e^{e^x}$ .

Solution:

Let $u = e^x$ , so $y = e^u$ .

\frac{dy}{dx} = e^u \cdot \frac{du}{dx} = e^{e^x} \cdot e^x = e^{x + e^x}

Example 4.6h: Nested Radicals

Problem: Differentiate $y = \sqrt{x + \sqrt{x}}$ .

Solution:

Write as $y = (x + x^{1/2})^{1/2}$ . Apply chain rule:

\frac{dy}{dx} = \frac{1}{2}(x + \sqrt{x})^{-1/2} \cdot \left(1 + \frac{1}{2\sqrt{x}}\right)

= \frac{1 + \frac{1}{2\sqrt{x}}}{2\sqrt{x + \sqrt{x}}} = \frac{2\sqrt{x} + 1}{4\sqrt{x}\sqrt{x + \sqrt{x}}}

Remark 4.1c: Recognizing Chain Rule Situations

You need the chain rule when you see:

Function of function: $\sin(x^2)$ , $e^{\cos x}$ , $\ln(1+x^2)$
Powers of expressions: $(3x+1)^5$ , $\sqrt{x^2+1}$
Trig of non-x: $\tan(3x)$ , $\cos(x^2)$
Exponentials: $e^{2x}$ , $2^{x^2}$
Logarithms: $\ln(x^3)$ , $\log(\sin x)$

Example 4.6i: Hyperbolic Functions with Chain Rule

Problem: Differentiate $y = \tanh(x^2)$ .

Solution:

Recall $(\tanh u)' = \text{sech}^2 u$ .

\frac{dy}{dx} = \text{sech}^2(x^2) \cdot 2x = 2x\,\text{sech}^2(x^2)

Example 4.6j: Related Rates Preview

Problem: If $A = \pi r^2$ and $r$ depends on time, find $\frac{dA}{dt}$ .

Solution:

Apply chain rule treating $r$ as $r(t)$ :

\frac{dA}{dt} = \frac{dA}{dr} \cdot \frac{dr}{dt} = 2\pi r \cdot \frac{dr}{dt}

This is the foundation for solving related rates problems in applications.

2. Inverse Function Derivatives

Theorem 4.5: Derivative of Inverse Function

If $f$ is differentiable and strictly monotonic on an interval, and $f'(x_0) \neq 0$ , then $f^{-1}$ is differentiable at $y_0 = f(x_0)$ with:

(f^{-1})'(y_0) = \frac{1}{f'(x_0)} = \frac{1}{f'(f^{-1}(y_0))}

Example 4.7: Derivative of arcsin x

Derive the formula for $(\arcsin x)'$ .

Let $y = \arcsin x$ , so $x = \sin y$ with $y \in (-\pi/2, \pi/2)$ .

\frac{dy}{dx} = \frac{1}{dx/dy} = \frac{1}{\cos y} = \frac{1}{\sqrt{1-\sin^2 y}} = \frac{1}{\sqrt{1-x^2}}

We use $\cos y > 0$ on $(-\pi/2, \pi/2)$ .

Proof of Theorem 4.5 (Inverse Function Derivative):

Let $y_0 = f(x_0)$ and suppose $f'(x_0) \neq 0$ .

(f^{-1})'(y_0) = \lim_{y \to y_0} \frac{f^{-1}(y) - f^{-1}(y_0)}{y - y_0}

Let $x = f^{-1}(y)$ , so $y = f(x)$ . As $y \to y_0$ , we have $x \to x_0$ :

= \lim_{x \to x_0} \frac{x - x_0}{f(x) - f(x_0)} = \frac{1}{\lim_{x \to x_0} \frac{f(x) - f(x_0)}{x - x_0}} = \frac{1}{f'(x_0)}

∎

Example 4.7b: Derivative of arccos x

Problem: Find $(\arccos x)'$ .

Solution:

Let $y = \arccos x$ , so $x = \cos y$ with $y \in (0, \pi)$ .

\frac{dy}{dx} = \frac{1}{dx/dy} = \frac{1}{-\sin y} = -\frac{1}{\sqrt{1-\cos^2 y}} = -\frac{1}{\sqrt{1-x^2}}

We use $\sin y > 0$ on $(0, \pi)$ .

Note: $(\arcsin x)' + (\arccos x)' = 0$ , confirming $\arcsin x + \arccos x = \frac{\pi}{2}$ .

Example 4.7c: Derivative of arctan x

Problem: Find $(\arctan x)'$ .

Solution:

Let $y = \arctan x$ , so $x = \tan y$ with $y \in (-\pi/2, \pi/2)$ .

\frac{dy}{dx} = \frac{1}{dx/dy} = \frac{1}{\sec^2 y} = \cos^2 y

Using $\sec^2 y = 1 + \tan^2 y = 1 + x^2$ :

(\arctan x)' = \frac{1}{1 + x^2}

Remark 4.2: Summary of Inverse Trigonometric Derivatives

(\arcsin x)' = \frac{1}{\sqrt{1-x^2}}

(\arccos x)' = -\frac{1}{\sqrt{1-x^2}}

(\arctan x)' = \frac{1}{1+x^2}

(\text{arccot } x)' = -\frac{1}{1+x^2}

(\text{arcsec } x)' = \frac{1}{|x|\sqrt{x^2-1}}

(\text{arccsc } x)' = -\frac{1}{|x|\sqrt{x^2-1}}

Example 4.7d: Chain Rule with Inverse Functions

Problem: Differentiate $y = \arctan(e^x)$ .

Solution:

Using chain rule with $u = e^x$ :

\frac{dy}{dx} = \frac{1}{1+(e^x)^2} \cdot e^x = \frac{e^x}{1+e^{2x}}

Example 4.7e: Inverse of a Specific Function

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

Solution:

First find $x_0$ such that $f(x_0) = 2$ .

$x^3 + x = 2$ gives $x_0 = 1$ (by inspection).

Since $f'(x) = 3x^2 + 1$ , we have $f'(1) = 4$ .

(f^{-1})'(2) = \frac{1}{f'(f^{-1}(2))} = \frac{1}{f'(1)} = \frac{1}{4}

3. Implicit Differentiation

When $y$ is defined implicitly by an equation $F(x, y) = 0$ , we differentiate both sides with respect to $x$ , treating $y$ as a function of $x$ .

Example 4.8: Circle

Find $\frac{dy}{dx}$ for the circle $x^2 + y^2 = r^2$ .

Differentiating implicitly:

2x + 2y\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{x}{y}

Example 4.9: Hyperbola

For $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ , find $y''$ .

First derivative: $\frac{2x}{a^2} - \frac{2y}{b^2}y' = 0$ , so $y' = \frac{b^2 x}{a^2 y}$

Second derivative (differentiating again):

y'' = \frac{b^2(y - xy')}{a^2 y^2} = \frac{b^2}{a^2 y^2}\left(y - \frac{b^2 x^2}{a^2 y}\right) = -\frac{b^4}{a^4 y^3}

Remark 4.3: Steps for Implicit Differentiation

Differentiate both sides with respect to $x$ , treating $y$ as a function of $x$ .
Apply chain rule when differentiating terms with $y$ : $\frac{d}{dx}[f(y)] = f'(y) \cdot \frac{dy}{dx}$ .
Solve for $\frac{dy}{dx}$ by collecting all terms with $\frac{dy}{dx}$ on one side.

Example 4.9b: Implicit Differentiation with Products

Problem: Find $\frac{dy}{dx}$ if $xy + y^2 = x^2$ .

Solution:

Differentiating implicitly:

\frac{d}{dx}(xy) + \frac{d}{dx}(y^2) = \frac{d}{dx}(x^2)

y + x\frac{dy}{dx} + 2y\frac{dy}{dx} = 2x

Solving for $\frac{dy}{dx}$ :

\frac{dy}{dx}(x + 2y) = 2x - y \implies \frac{dy}{dx} = \frac{2x - y}{x + 2y}

Example 4.9c: Implicit Differentiation with Trigonometric Functions

Problem: Find $\frac{dy}{dx}$ if $\sin(xy) = x$ .

Solution:

Differentiating using chain rule on left side:

\cos(xy) \cdot \frac{d}{dx}(xy) = 1

\cos(xy) \cdot \left(y + x\frac{dy}{dx}\right) = 1

Solving for $\frac{dy}{dx}$ :

x\cos(xy)\frac{dy}{dx} = 1 - y\cos(xy)

\frac{dy}{dx} = \frac{1 - y\cos(xy)}{x\cos(xy)}

Example 4.9d: Finding Tangent Line via Implicit Differentiation

Problem: Find the equation of the tangent line to $x^3 + y^3 = 6xy$ at $(3, 3)$ .

Solution:

Verify point: $27 + 27 = 54 = 6(9)$ ✓

Differentiating implicitly:

3x^2 + 3y^2\frac{dy}{dx} = 6y + 6x\frac{dy}{dx}

\frac{dy}{dx}(3y^2 - 6x) = 6y - 3x^2

\frac{dy}{dx} = \frac{6y - 3x^2}{3y^2 - 6x} = \frac{2y - x^2}{y^2 - 2x}

At $(3, 3)$ :

\frac{dy}{dx}\Big|_{(3,3)} = \frac{6 - 9}{9 - 6} = \frac{-3}{3} = -1

Tangent line: $y - 3 = -1(x - 3)$ , i.e., $y = -x + 6$ .

Remark 4.3b: Logarithmic Differentiation

For functions of the form $y = [f(x)]^{g(x)}$ or products of many factors, use logarithmic differentiation:

Take $\ln$ of both sides: $\ln y = g(x) \ln f(x)$
Differentiate implicitly: $\frac{1}{y}\frac{dy}{dx} = \ldots$
Multiply both sides by $y$ to get $\frac{dy}{dx}$

Example 4.9e: Logarithmic Differentiation

Problem: Differentiate $y = x^{\sin x}$ for $x > 0$ .

Solution:

Take ln of both sides:

\ln y = \sin x \cdot \ln x

Differentiate implicitly:

\frac{1}{y}\frac{dy}{dx} = \cos x \cdot \ln x + \sin x \cdot \frac{1}{x}

Multiply by $y = x^{\sin x}$ :

\frac{dy}{dx} = x^{\sin x}\left(\cos x \ln x + \frac{\sin x}{x}\right)

4. Parametric Derivatives

Theorem 4.6: Derivatives of Parametric Curves

For a curve given by $x = \phi(t), y = \psi(t)$ :

\frac{dy}{dx} = \frac{\psi'(t)}{\phi'(t)} = \frac{dy/dt}{dx/dt}

For the second derivative:

\frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{dy}{dx}\right) \cdot \frac{1}{dx/dt} = \frac{\psi''\phi' - \psi'\phi''}{(\phi')^3}

Example 4.10: Cycloid

For the cycloid $x = t - \sin t, y = 1 - \cos t$ , find $y''$ .

\frac{dx}{dt} = 1 - \cos t, \quad \frac{dy}{dt} = \sin t

\frac{dy}{dx} = \frac{\sin t}{1 - \cos t}

For the second derivative, differentiate and divide by $dx/dt$ :

\frac{d^2y}{dx^2} = -\frac{1}{(1-\cos t)^2}

Proof of Theorem 4.6 (Parametric Derivatives):

Since $y$ is a function of $t$ and $t$ is implicitly a function of $x$ :

\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx}

Using $\frac{dt}{dx} = \frac{1}{dx/dt}$ (inverse function derivative):

\frac{dy}{dx} = \frac{dy/dt}{dx/dt}

For the second derivative:

\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dt}\left(\frac{dy}{dx}\right) \cdot \frac{dt}{dx}

∎

Example 4.10b: Ellipse in Parametric Form

Problem: For the ellipse $x = a\cos t, y = b\sin t$ , find $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$ .

Solution:

\frac{dx}{dt} = -a\sin t, \quad \frac{dy}{dt} = b\cos t

\frac{dy}{dx} = \frac{b\cos t}{-a\sin t} = -\frac{b}{a}\cot t

For the second derivative:

\frac{d}{dt}\left(\frac{dy}{dx}\right) = -\frac{b}{a} \cdot (-\csc^2 t) = \frac{b}{a}\csc^2 t

\frac{d^2y}{dx^2} = \frac{(b/a)\csc^2 t}{-a\sin t} = -\frac{b}{a^2}\csc^3 t = -\frac{b}{a^2\sin^3 t}

Example 4.10c: Parabola in Parametric Form

Problem: For $x = t^2, y = t^3$ , find the tangent line at $t = 2$ .

Solution:

At $t = 2$ : $x = 4$ , $y = 8$ .

\frac{dy}{dx} = \frac{3t^2}{2t} = \frac{3t}{2}

At $t = 2$ : slope = $\frac{3(2)}{2} = 3$ .

Tangent line: $y - 8 = 3(x - 4)$ , i.e., $y = 3x - 4$ .

Remark 4.4: Polar Coordinates

For a curve in polar coordinates $r = f(\theta)$ , we can convert to parametric form:

x = r\cos\theta = f(\theta)\cos\theta, \quad y = r\sin\theta = f(\theta)\sin\theta

Then apply parametric derivative formulas:

\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{f'(\theta)\sin\theta + f(\theta)\cos\theta}{f'(\theta)\cos\theta - f(\theta)\sin\theta}

Example 4.10d: Polar Curve Derivative

Problem: For the cardioid $r = 1 + \cos\theta$ , find $\frac{dy}{dx}$ at $\theta = \pi/2$ .

Solution:

We have $r = 1 + \cos\theta$ , $r' = -\sin\theta$ .

\frac{dy}{dx} = \frac{r'\sin\theta + r\cos\theta}{r'\cos\theta - r\sin\theta}

At $\theta = \pi/2$ : $r = 1$ , $r' = -1$ , $\sin\theta = 1$ , $\cos\theta = 0$ .

\frac{dy}{dx} = \frac{(-1)(1) + (1)(0)}{(-1)(0) - (1)(1)} = \frac{-1}{-1} = 1

5. Higher-Order Derivatives

Definition 4.3: Higher-Order Derivatives

The nth derivative of $f$ is defined recursively:

f^{(n)}(x) = \frac{d}{dx}f^{(n-1)}(x), \quad f^{(0)}(x) = f(x)

Notation: $f'', f''', f^{(4)}, \ldots$ or $\frac{d^2y}{dx^2}, \frac{d^3y}{dx^3}, \ldots$

Standard Higher Derivatives

(e^{ax})^{(n)} = a^n e^{ax}

(\sin x)^{(n)} = \sin\left(x + \frac{n\pi}{2}\right)

(\cos x)^{(n)} = \cos\left(x + \frac{n\pi}{2}\right)

(x^m)^{(n)} = \frac{m!}{(m-n)!}x^{m-n} \text{ for } n \leq m

Theorem 4.7: Leibniz's Formula

For the nth derivative of a product:

(fg)^{(n)} = \sum_{k=0}^{n} \binom{n}{k} f^{(k)} g^{(n-k)}

This generalizes the product rule to arbitrary order.

Example 4.11: Leibniz Formula

Find $(x^2 e^x)^{(n)}$ .

(x^2 e^x)^{(n)} = \sum_{k=0}^{\min(n,2)} \binom{n}{k} (x^2)^{(k)} (e^x)^{(n-k)}

= \binom{n}{0}x^2 e^x + \binom{n}{1}(2x)e^x + \binom{n}{2}(2)e^x

= e^x\left(x^2 + 2nx + n(n-1)\right)

Proof of Theorem 4.7 (Leibniz's Formula):

By induction on $n$ :

Base case: $n = 1$ is the product rule: $(fg)' = f'g + fg'$ .

Inductive step: Assume true for $n$ :

(fg)^{(n+1)} = \frac{d}{dx}\left[\sum_{k=0}^{n} \binom{n}{k} f^{(k)} g^{(n-k)}\right]

= \sum_{k=0}^{n} \binom{n}{k} \left[f^{(k+1)} g^{(n-k)} + f^{(k)} g^{(n+1-k)}\right]

Reindexing and using $\binom{n}{k} + \binom{n}{k-1} = \binom{n+1}{k}$ :

= \sum_{k=0}^{n+1} \binom{n+1}{k} f^{(k)} g^{(n+1-k)}

∎

Example 4.11b: nth Derivative of Polynomial × Exponential

Problem: Find $(x^3 e^{2x})^{(n)}$ .

Solution:

Using Leibniz formula with $f = x^3$ , $g = e^{2x}$ :

$(x^3)^{(0)} = x^3$ , $(x^3)^{(1)} = 3x^2$ , $(x^3)^{(2)} = 6x$ , $(x^3)^{(3)} = 6$
$(e^{2x})^{(k)} = 2^k e^{2x}$

(x^3 e^{2x})^{(n)} = \sum_{k=0}^{\min(n,3)} \binom{n}{k} (x^3)^{(k)} 2^{n-k} e^{2x}

For $n \geq 3$ :

= 2^n e^{2x}\left[x^3 + \frac{3n}{2}x^2 + \frac{3n(n-1)}{4}x + \frac{n(n-1)(n-2)}{8}\right]

Example 4.11c: nth Derivative of Trigonometric Functions

Problem: Find $(x\sin x)^{(n)}$ .

Solution:

Using Leibniz formula with $f = x$ , $g = \sin x$ :

(x\sin x)^{(n)} = \binom{n}{0}x \cdot \sin^{(n)}x + \binom{n}{1} \cdot 1 \cdot \sin^{(n-1)}x

Using $\sin^{(k)}x = \sin(x + k\pi/2)$ :

= x\sin\left(x + \frac{n\pi}{2}\right) + n\sin\left(x + \frac{(n-1)\pi}{2}\right)

Remark 4.5: Higher Derivatives of Standard Functions

Function	nth Derivative
$e^{ax}$	$a^n e^{ax}$
$\sin(ax)$	$a^n \sin(ax + n\pi/2)$
$\cos(ax)$	$a^n \cos(ax + n\pi/2)$
$\ln x$	$\frac{(-1)^{n-1}(n-1)!}{x^n}$ for $n \geq 1$
$x^m$ ( $n \leq m$ )	$\frac{m!}{(m-n)!}x^{m-n}$
$\frac{1}{x-a}$	$\frac{(-1)^n n!}{(x-a)^{n+1}}$

Example 4.11d: nth Derivative of Rational Function

Problem: Find $\left(\frac{1}{1-x}\right)^{(n)}$ .

Solution:

Write as $(1-x)^{-1}$ and differentiate:

$f^{(1)} = (1-x)^{-2}$
$f^{(2)} = 2(1-x)^{-3}$
$f^{(3)} = 6(1-x)^{-4}$

Pattern: $f^{(n)} = n!(1-x)^{-(n+1)} = \frac{n!}{(1-x)^{n+1}}$

Definition 4.4: Higher-Order Differentials

The nth differential of $y = f(x)$ is defined as:

d^n y = f^{(n)}(x) \, (dx)^n

Note: This defines $d^2y = f''(x)(dx)^2$ , $d^3y = f'''(x)(dx)^3$ , etc.

Remark 4.5b: Physical Interpretation

Higher derivatives have important physical meanings:

Position: $s(t)$

Describes where an object is at time $t$

Velocity: $v = s'(t)$

Rate of change of position

Acceleration: $a = s''(t)$

Rate of change of velocity

Jerk: $j = s'''(t)$

Rate of change of acceleration

Example 4.11e: Curvature Formula

Connection: The curvature of a curve $y = f(x)$ involves the second derivative:

\kappa = \frac{|f''(x)|}{(1 + [f'(x)]^2)^{3/2}}

Curvature measures how sharply a curve bends. It uses both first and second derivatives!

Remark 4.5c: C^n Functions

We classify functions by their differentiability:

C⁰Continuous functions
C¹Continuously differentiable (f' exists and is continuous)
C²f, f', f'' all continuous
CⁿFirst n derivatives exist and are continuous
C^∞Smooth functions: infinitely differentiable (e.g., sin, cos, e^x)

Example 4.11f: Function That Is Not C^∞

Example: Show $f(x) = |x|^3$ is $C^2$ but not $C^3$ at $x = 0$ .

Solution:

For $x \neq 0$ :

$f(x) = |x|^3 = \begin{cases} x^3 & x > 0 \\ -x^3 & x < 0 \end{cases}$
$f'(x) = 3|x| \cdot \text{sgn}(x) \cdot |x| = 3x|x|$
$f''(x) = 6|x|$

At $x = 0$ : $f''(0) = 0$ and $f''$ is continuous.

But $f'''(x)$ jumps from $-6$ to $+6$ at $x = 0$ .

Conclusion: $f \in C^2$ but $f \notin C^3$ at $x = 0$ .

Corollary 4.1: Pattern for nth Derivative

Many functions have elegant closed-form nth derivatives:

Polynomial $x^m$ :

(x^m)^{(n)} = \begin{cases} \frac{m!}{(m-n)!}x^{m-n} & n \leq m \\ 0 & n > m \end{cases}

Shifted power $(x-a)^m$ :

\left[(x-a)^m\right]^{(n)} = \frac{m!}{(m-n)!}(x-a)^{m-n}

Partial fractions:

\left(\frac{1}{x-a}\right)^{(n)} = \frac{(-1)^n n!}{(x-a)^{n+1}}

Chapter Summary

This module covered essential techniques for differentiating complex functions:

1.Chain Rule allows differentiation of composite functions by multiplying the derivative of the outer function by the derivative of the inner function.
2.Inverse Function Derivatives follow the formula $(f^{-1})'(y) = 1/f'(x)$ , leading to derivatives of arcsin, arccos, arctan, etc.
3.Implicit Differentiation handles equations where y is not explicitly solved, by treating y as y(x) and applying chain rule.
4.Parametric Derivatives use $dy/dx = (dy/dt)/(dx/dt)$ for curves given parametrically.
5.Higher Derivatives and Leibniz's formula extend differentiation to arbitrary orders.

Chain Rule & Higher Derivatives Practice

Questions

Correct

Accuracy

y = \sin(x^2)

, what is

\frac{dy}{dx}

Easy

Not attempted

y = e^{\sin x}

, find

\frac{dy}{dx}

Easy

Not attempted

For the inverse function, if

f'(x) = 3x^2

and

f(2) = 8

, what is

(f^{-1})'(8)

Medium

Not attempted

Using implicit differentiation on

x^2 + y^2 = 25

, find

\frac{dy}{dx}

Easy

Not attempted

x = t^2

and

y = t^3

, what is

\frac{dy}{dx}

Medium

Not attempted

What is the second derivative of

f(x) = e^{2x}

Easy

Not attempted

Find

\frac{d^2y}{dx^2}

for the parametric curve

x = \cos t

y = \sin t

Hard

Not attempted

What is the nth derivative of

f(x) = e^{ax}

Medium

Not attempted

Find

(\ln(x^2+1))'

Easy

Not attempted

y = \arctan x

, what is

y'

Easy

Not attempted

For

e^y = xy

, find

\frac{dy}{dx}

using implicit differentiation.

Medium

Not attempted

What is

(\sin^2 x)'

Easy

Not attempted

Find

(x^x)'

for

x > 0

Hard

Not attempted

For

x = t^3

y = t^2

, find

\frac{d^2y}{dx^2}

Hard

Not attempted

Using Leibniz formula,

(xe^x)'' =

Medium

Not attempted

Frequently Asked Questions

Why is the chain rule important?

The chain rule allows us to differentiate composite functions, which appear everywhere in applications. Without it, we couldn't differentiate functions like sin(x²), e^(cos x), or ln(1+x²).

How do I know when to use implicit differentiation?

Use implicit differentiation when y is not explicitly given as a function of x, such as in equations like x² + y² = 1 or xy + sin(y) = x.

What's the connection between parametric derivatives and chain rule?

The formula dy/dx = (dy/dt)/(dx/dt) comes from the chain rule: dy/dx = (dy/dt)·(dt/dx) = (dy/dt)/(dx/dt).

Is d²y/dx² the same as (dy/dx)²?

No! d²y/dx² is the second derivative (derivative of the derivative), while (dy/dx)² is the square of the first derivative. These are different quantities.

How do I identify the 'inner' and 'outer' functions?

The outer function is applied last. In f(g(x)), f is outer, g is inner. Example: sin(x²) has outer=sin, inner=x². Think about order of operations.

When should I use logarithmic differentiation?

Use it for: (1) y = f(x)^g(x) where both base and exponent contain x, (2) products of many terms, (3) complicated quotients. It converts products to sums.

How do I find the second derivative for parametric curves?

First find dy/dx = (dy/dt)/(dx/dt). Then d²y/dx² = (d/dt)(dy/dx) ÷ (dx/dt). Don't just differentiate dy/dx directly with respect to t!

What's the pattern for nth derivative of e^(ax)?

Each differentiation brings down a factor of a: (e^(ax))^(n) = a^n · e^(ax). This is because d/dx(e^(ax)) = a·e^(ax).

How do I remember Leibniz's formula?

It's like the binomial theorem: (f·g)^(n) = Σ C(n,k) f^(k) g^(n-k). The binomial coefficients distribute the n derivatives between f and g.

What happens to inverse trig derivatives with chain rule?

Apply chain rule as usual. Example: d/dx[arctan(2x)] = 1/(1+(2x)²) · 2 = 2/(1+4x²). The inner function derivative multiplies.

Key Takeaways

Chain Rule

(f \circ g)'(x) = f'(g(x)) \cdot g'(x)

Inverse Function

(f^{-1})'(y) = \frac{1}{f'(f^{-1}(y))}

Parametric Derivative

\frac{dy}{dx} = \frac{dy/dt}{dx/dt}

Leibniz Formula

(fg)^{(n)} = \sum_{k=0}^{n} \binom{n}{k} f^{(k)} g^{(n-k)}

Implicit Differentiation

Treat y as y(x), apply chain rule

Logarithmic Diff.

Take ln, differentiate, multiply by y

arcsin x

(\arcsin x)' = \frac{1}{\sqrt{1-x^2}}

arctan x

(\arctan x)' = \frac{1}{1+x^2}

Study Tips

Chain Rule Strategy

Work from outside in. Identify the outermost function first, differentiate it, then multiply by the derivative of the inner function.

Implicit Differentiation

Always remember to write dy/dx whenever you differentiate a term containing y. Then solve for dy/dx algebraically.

Parametric Curves

For second derivatives, don't forget to divide by dx/dt again. Common mistake: differentiating dy/dx with respect to t only.

Higher Derivatives

Look for patterns! Many functions have predictable nth derivative formulas. Memorize key patterns for exponentials, trig, and powers.

What's Next?

In the next module, you will learn about the Mean Value Theorems, fundamental results that connect derivatives to function behavior.

Rolle's Theorem: zeros of derivatives between zeros of functions
Lagrange's Mean Value Theorem: $f(b)-f(a) = f'(c)(b-a)$
Cauchy's Mean Value Theorem: generalization for two functions
Applications to proving inequalities and limits