Derivative basics

This chapter establishes the foundational concepts of derivatives, starting with the formal definition and progressing through essential computational techniques, including standard formulas and composite function rules. A thorough understanding of these principles is critical for success in subsequent calculus topics and is frequently assessed in CMI examinations.

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Chapter Contents

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| Topic |

|---|-------| | 1 | Derivative from first principles | | 2 | Standard derivative formulas | | 3 | One-sided derivatives | | 4 | Derivative of composite functions |

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We begin with Derivative from first principles.

Part 1: Derivative from first principles

Derivative from First Principles

Overview

The derivative from first principles is the foundational definition of differentiation. Before using shortcut rules like the power rule or product rule, we define the derivative as the limiting value of the average rate of change. In CMI-style questions, this topic tests algebraic simplification, limit handling, and geometric understanding together. ---

Learning Objectives

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Core Idea

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Meaning of the Formula

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Equivalent Forms

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Standard First-Principles Computations

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Minimal Worked Examples

Example 1 Find the derivative of $f(x)=x^2$ from first principles. We use $\qquad f'(x)=\lim_{h\to 0}\dfrac{f(x+h)-f(x)}{h}$ Now, $\qquad f(x+h)=(x+h)^2=x^2+2xh+h^2$ So, $\qquad \dfrac{f(x+h)-f(x)}{h}=\dfrac{x^2+2xh+h^2-x^2}{h}$ $\qquad =\dfrac{2xh+h^2}{h}=2x+h$ Taking limit as $h\to 0$, $\qquad f'(x)=\lim_{h\to 0}(2x+h)=2x$ So the derivative is $\boxed{2x}$. --- Example 2 Find the derivative of $f(x)=\dfrac{1}{x}$ from first principles. We compute $\qquad f'(x)=\lim_{h\to 0}\dfrac{\dfrac{1}{x+h}-\dfrac{1}{x}}{h}$ Combine the fractions: $\qquad \dfrac{1}{x+h}-\dfrac{1}{x}=\dfrac{x-(x+h)}{x(x+h)}=\dfrac{-h}{x(x+h)}$ So, $\qquad \dfrac{f(x+h)-f(x)}{h}=\dfrac{-h}{x(x+h)}\cdot \dfrac{1}{h}=-\dfrac{1}{x(x+h)}$ Taking limit as $h\to 0$, $\qquad f'(x)=-\dfrac{1}{x^2}$ So the derivative is $\boxed{-\dfrac{1}{x^2}}$. ---

Standard Results That Come from First Principles

But in this topic, the goal is to understand where these rules come from, not just to use them. ---

When the Derivative May Not Exist

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Right-Hand and Left-Hand Derivatives

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Common Algebra Required

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Common Mistakes

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="Using first principles, the derivative of $f(x)=x^2$ at $x=1$ is" options=["$1$","$2$","$3$","$4$"] answer="B" hint="Compute $f'(1)$ from the definition or use the known result after deriving it." solution="Using first principles, $\qquad f'(1)=\lim_{h\to 0}\dfrac{(1+h)^2-1}{h}$ Expand: $\qquad (1+h)^2=1+2h+h^2$ So, $\qquad \dfrac{(1+h)^2-1}{h}=\dfrac{2h+h^2}{h}=2+h$ Taking the limit as $h\to 0$, we get $\qquad f'(1)=2$ Hence the correct option is $\boxed{B}$." ::: :::question type="NAT" question="Find $f'(2)$ from first principles for $f(x)=x^2-3x+1$." answer="1" hint="Use $f'(a)=\lim_{h\to 0}\dfrac{f(a+h)-f(a)}{h}$ with $a=2$." solution="We compute $\qquad f'(2)=\lim_{h\to 0}\dfrac{f(2+h)-f(2)}{h}$ Now, $\qquad f(2+h)=(2+h)^2-3(2+h)+1$ $\qquad =4+4h+h^2-6-3h+1$ $\qquad =h^2+h-1$ Also, $\qquad f(2)=2^2-3\cdot 2+1=4-6+1=-1$ So, $\qquad \dfrac{f(2+h)-f(2)}{h}=\dfrac{(h^2+h-1)-(-1)}{h}$ $\qquad =\dfrac{h^2+h}{h}=h+1$ Taking limit as $h\to 0$, $\qquad f'(2)=1$ Therefore the answer is $\boxed{1}$." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["$f'(a)=\lim_{h\to0}\dfrac{f(a+h)-f(a)}{h}$, if the limit exists","The derivative at a point represents the slope of the tangent there","A function can be differentiable at a point without being continuous there","For $f(x)=|x|$, the derivative at $x=0$ does not exist"] answer="A,B,D" hint="Recall the definition and the continuity-differentiability relation." solution="1. True. This is the first-principles definition.

True. The derivative gives the slope of the tangent line.

False. Differentiability implies continuity.

True. For $|x|$, left and right derivatives at $0$ are different.

Hence the correct answer is $\boxed{A,B,D}$." ::: :::question type="SUB" question="Using first principles, find the derivative of $f(x)=2x+5$." answer="$2$" hint="Start from $\dfrac{f(x+h)-f(x)}{h}$." solution="We use $\qquad f'(x)=\lim_{h\to 0}\dfrac{f(x+h)-f(x)}{h}$ Now, $\qquad f(x+h)=2(x+h)+5=2x+2h+5$ So, $\qquad \dfrac{f(x+h)-f(x)}{h}=\dfrac{(2x+2h+5)-(2x+5)}{h}$ $\qquad =\dfrac{2h}{h}=2$ Taking the limit, $\qquad f'(x)=2$ Hence the derivative is $\boxed{2}$." ::: ---

Summary

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Part 2: Standard derivative formulas

Standard Derivative Formulas

Overview

Standard derivative formulas are the basic tools used in almost every differentiation problem. In CMI-style questions, these formulas are rarely tested in isolation; instead, they appear inside simplification, tangent problems, monotonicity, rate of change, and approximation arguments. So the real goal is not just memorisation, but fast and correct recognition. ---

Learning Objectives

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Core Idea

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Most Important Standard Formulas

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Supporting Rules Used with Standard Formulas

These rules are what make standard formulas actually useful in real questions. ---

High-Value Derived Forms

These come from combining the standard formulas with the chain rule. ---

Domain and Validity

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Common Mistakes

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Minimal Worked Examples

Example 1 Differentiate $f(x)=x^5-3x^2+7$. Using standard algebraic formulas, $\qquad f'(x)=5x^4-6x$ --- Example 2 Differentiate $g(x)=\sin(3x)$. Using chain rule, $\qquad g'(x)=\cos(3x)\cdot 3=3\cos(3x)$ ---

Formula Table for Fast Revision

| Function | Derivative | |---|---| | $c$ | $0$ | | $x^n$ | $n x^{n-1}$ | | $\dfrac{1}{x}$ | $-\dfrac{1}{x^2}$ | | $\sqrt{x}$ | $\dfrac{1}{2\sqrt{x}}$ | | $\sin x$ | $\cos x$ | | $\cos x$ | $-\sin x$ | | $\tan x$ | $\sec^2 x$ | | $e^x$ | $e^x$ | | $a^x$ | $a^x\ln a$ | | $\ln x$ | $\dfrac{1}{x}$ | ---

CMI Strategy

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Practice Questions

:::question type="MCQ" question="The derivative of $\cos x$ is" options=["$\sin x$","$-\sin x$","$\cos x$","$-\cos x$"] answer="B" hint="This is one of the most common sign traps." solution="The standard derivative formula is $\qquad \dfrac{d}{dx}(\cos x)=-\sin x$. Hence the correct option is $\boxed{B}$." ::: :::question type="NAT" question="Find the value of $\dfrac{d}{dx}(x^4-2x^2+5)$ at $x=2$." answer="24" hint="Differentiate first, then substitute $x=2$." solution="Differentiate: $\qquad \dfrac{d}{dx}(x^4-2x^2+5)=4x^3-4x$ Now substitute $x=2$: $\qquad 4(2^3)-4(2)=32-8=24$ Therefore the answer is $\boxed{24}$." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["$\dfrac{d}{dx}(e^x)=e^x$","$\dfrac{d}{dx}(\ln x)=\dfrac{1}{x}$ for $x>0$","$\dfrac{d}{dx}(\sin x)=-\cos x$","$\dfrac{d}{dx}(x^n)=n x^{n-1}$"] answer="A,B,D" hint="Check each standard formula carefully." solution="1. True.

True for $x>0$.

False, because $\dfrac{d}{dx}(\sin x)=\cos x$.

True.

Hence the correct answer is $\boxed{A,B,D}$." ::: :::question type="SUB" question="Differentiate $f(x)=x^3+\dfrac{1}{x}-\ln x$." answer="$3x^2-\dfrac{1}{x^2}-\dfrac{1}{x}$" hint="Differentiate term by term." solution="Using standard formulas term by term: $\qquad \dfrac{d}{dx}(x^3)=3x^2$ $\qquad \dfrac{d}{dx}\left(\dfrac{1}{x}\right)=-\dfrac{1}{x^2}$ $\qquad \dfrac{d}{dx}(\ln x)=\dfrac{1}{x}$ So, $\qquad f'(x)=3x^2-\dfrac{1}{x^2}-\dfrac{1}{x}$ Hence the derivative is $\boxed{3x^2-\dfrac{1}{x^2}-\dfrac{1}{x}}$." ::: ---

Summary

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Part 3: One-sided derivatives

One-sided Derivatives

Overview

One-sided derivatives are local rate-of-change limits taken from only one side of a point. They become important when:

• the function is piecewise-defined,

• the point is a boundary point,

• the left and right behaviour are different.

In CMI-style questions, one-sided derivatives are often used to test whether a function is differentiable, whether continuity follows, and how to handle functions like $|x|$ or piecewise formulas. ---

Learning Objectives

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Definitions

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Immediate Consequences

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Right derivative implies right continuity

Why? If $\qquad \dfrac{f(a+h)-f(a)}{h}\to L$ as $h\to 0^+$, then $\qquad f(a+h)-f(a)=h\cdot \dfrac{f(a+h)-f(a)}{h}\to 0$ as $h\to 0^+$. So the function values from the right approach $f(a)$. ---

Most Common Examples

1. Smooth function

For $f(x)=x^2$ at any point $a$, $\qquad f'_+(a)=f'_-(a)=2a$ So $f$ is differentiable everywhere. ---

2. Absolute value at the origin

For $f(x)=|x|$ at $x=0$, Right derivative: $\qquad f'_+(0)=\lim_{h\to 0^+}\dfrac{|h|-0}{h}=\lim_{h\to 0^+}\dfrac{h}{h}=1$ Left derivative: $\qquad f'_-(0)=\lim_{h\to 0^-}\dfrac{|h|-0}{h}=\lim_{h\to 0^-}\dfrac{-h}{h}=-1$ Since the one-sided derivatives are unequal, $|x|$ is not differentiable at $0$. But it does have both one-sided derivatives. ---

3. Continuous but no finite right derivative

Consider $f(x)=\sqrt{x}$ at $x=0$ on $[0,\infty)$. It is continuous at $0$, but $\qquad f'_+(0)=\lim_{h\to 0^+}\dfrac{\sqrt{h}-0}{h}=\lim_{h\to 0^+}\dfrac{1}{\sqrt{h}}$ which is not finite. So continuity does not force existence of a finite right derivative. ---

Piecewise Functions

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Endpoint view

This is why one-sided derivatives are not just technical details; they are natural for boundary points. ---

Minimal Worked Examples

Example 1 Let $\qquad f(x)= \begin{cases} 0, & x<0 \\ x, & x\ge 0 \end{cases} $ at $x=0$. Right derivative: $\qquad f'_+(0)=\lim_{h\to 0^+}\dfrac{h-0}{h}=1$ Left derivative: $\qquad f'_-(0)=\lim_{h\to 0^-}\dfrac{0-0}{h}=0$ So $f$ is continuous at $0$, but not differentiable there. --- Example 2 Let $\qquad f(x)= \begin{cases} 0, & x\le 1 \\ (x-1)^2, & x>1 \end{cases} $ Find the right derivative at $x=1$. $\qquad f'_+(1)=\lim_{h\to 0^+}\dfrac{(1+h-1)^2-0}{h}$ $\qquad = \lim_{h\to 0^+}\dfrac{h^2}{h}=\lim_{h\to 0^+}h=0$ So the right derivative is $\boxed{0}$. ---

Common Mistakes

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="For the function $f(x)=|x|$ at $x=0$, which statement is correct?" options=["Both one-sided derivatives are equal","The right derivative exists and equals $1$","The left derivative exists and equals $1$","The function is differentiable at $0$"] answer="B" hint="Compute from the right and from the left separately." solution="For $f(x)=|x|$, $\qquad f'_+(0)=1$ and $\qquad f'_-(0)=-1$ So the one-sided derivatives are not equal. Hence $f$ is not differentiable at $0$. Therefore the correct option is $\boxed{B}$." ::: :::question type="NAT" question="Find the right derivative of $f(x)=|x|$ at $x=0$." answer="1" hint="For $h>0$, $|h|=h$." solution="By definition, $\qquad f'_+(0)=\lim_{h\to 0^+}\dfrac{|h|-0}{h}$ Since $h>0$, we have $|h|=h$. So $\qquad f'_+(0)=\lim_{h\to 0^+}\dfrac{h}{h}=1$ Hence the answer is $\boxed{1}$." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["If $f$ is differentiable at $x=a$, then $f$ has a right derivative at $x=a$","$f(x)=|x|$ has a right derivative at $x=0$","If $f$ has a right derivative at $x=a$, then $f$ must be continuous at $x=a$","If $f$ is continuous at $x=a$, then $f$ must have a right derivative at $x=a$"] answer="A,B" hint="Think carefully about right continuity versus full continuity." solution="1. True. Differentiability implies existence of both one-sided derivatives.

True. For $|x|$ at $0$, the right derivative is $1$.

False. A right derivative implies right continuity, not necessarily full continuity.

False. Continuity alone does not guarantee existence of a finite right derivative. For example, $\sqrt{x}$ is continuous at $0$ on $[0,\infty)$, but its right derivative at $0$ is not finite. Hence the correct answer is $\boxed{A,B}$." ::: :::question type="SUB" question="Let $\qquad f(x)= \begin{cases} 0, & x<0 \\ x, & x\ge 0 \end{cases} $ Find the left and right derivatives at $x=0$, and determine whether $f$ is differentiable at $0$." answer="Left derivative $=0$, right derivative $=1$, so not differentiable at $0$" hint="Use the branch valid on each side of $0$." solution="For $h>0$, $\qquad f(h)=h,\quad f(0)=0$ So the right derivative is $\qquad f'_+(0)=\lim_{h\to 0^+}\dfrac{h-0}{h}=1$ For $h<0$, $\qquad f(h)=0,\quad f(0)=0$ So the left derivative is $\qquad f'_-(0)=\lim_{h\to 0^-}\dfrac{0-0}{h}=0$ Since $\qquad f'_-(0)\ne f'_+(0)$, the function is not differentiable at $0$. Thus the result is $\qquad \boxed{f'_-(0)=0,\ f'_+(0)=1,\ \text{not differentiable at }0}$." ::: ---

Summary

❗ Key Takeaways for CMI

• One-sided derivatives are limits taken from only one side of a point.

• Differentiability at an interior point means both one-sided derivatives exist and are equal.

• $|x|$ at $0$ is the standard example where one-sided derivatives exist but are unequal.

• A right derivative gives right continuity, not necessarily full continuity.

• Continuity alone does not guarantee existence of a finite one-sided derivative.

• Piecewise and endpoint questions are the natural home of one-sided derivatives.

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💡 Next Up

Proceeding to Derivative of composite functions.

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Part 4: Derivative of composite functions

Derivative of Composite Functions

Overview

A composite function is a function inside another function. In differentiation, this topic is governed by the chain rule, one of the most important rules in calculus. In CMI-style questions, the chain rule is often not tested in a direct “differentiate this” way. Instead, it appears inside related rates, parametrised motion, arc-length problems, implicit dependence, and expressions where one quantity depends on another which itself depends on time or another variable. ---

Learning Objectives

❗ By the End of This Topic

After studying this topic, you will be able to:

• Recognise composite functions correctly.

• Apply the chain rule to two-layer and multi-layer compositions.

• Differentiate expressions involving powers, radicals, exponentials, logarithms, and trigonometric composites.

• Handle related-rate situations where one variable depends on another which depends on time.

• Convert geometric or motion-based information into a chain-rule calculation.

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Core Idea

📖 Composite Function

If
$\qquad y = f(g(x))$
then $y$ is called a composite function.

Here:

• $g(x)$ is the inner function

• $f$ acts on the output of $g(x)$, so $f$ is the outer function

📐 Chain Rule

If
$\qquad y=f(g(x))$
then

$\qquad \dfrac{dy}{dx} = f'(g(x))\cdot g'(x)$

This means:

differentiate the outer function

keep the inner part unchanged

multiply by the derivative of the inner part

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Why the Chain Rule Matters

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Standard Chain-Rule Forms

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Two-Layer and Multi-Layer Composition

Each layer contributes a derivative factor. For example, if $\qquad y=\sin\big((x^2+1)^3\big)$ then the layers are:

• outer: $\sin(\cdot)$

• middle: $(\cdot)^3$

• inner: $x^2+1$

So $\qquad \dfrac{dy}{dx}=\cos\big((x^2+1)^3\big)\cdot 3(x^2+1)^2\cdot 2x$ ::: ---

Minimal Worked Examples

Example 1 Differentiate $\qquad y=(2x+1)^7$ Using chain rule, $\qquad \dfrac{dy}{dx}=7(2x+1)^6\cdot 2$ $\qquad = 14(2x+1)^6$ --- Example 2 Differentiate $\qquad y=\ln(1+x^2)$ Outer function is $\ln(\cdot)$ and inner function is $1+x^2$. So, $\qquad \dfrac{dy}{dx}=\dfrac{1}{1+x^2}\cdot 2x$ $\qquad = \dfrac{2x}{1+x^2}$ ---

Composite Functions in Related Rates

This is one of the most important applications of the chain rule. ---

PYQ-Type Insight: Spider Problem Structure

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Important Formula Patterns for This Topic

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How to Identify the Inner Function

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Common Mistakes

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="The derivative of $(x^2+1)^4$ is" options=["$4(x^2+1)^3$","$8x(x^2+1)^3$","$4x(x^2+1)^3$","$8(x^2+1)^3$"] answer="B" hint="Use chain rule with outer function $(\cdot)^4$." solution="Let $\qquad y=(x^2+1)^4$ Then $\qquad \dfrac{dy}{dx}=4(x^2+1)^3\cdot \dfrac{d}{dx}(x^2+1)$ $\qquad = 4(x^2+1)^3\cdot 2x = 8x(x^2+1)^3$ Hence the correct option is $\boxed{B}$." ::: :::question type="NAT" question="Find the value of $\dfrac{d}{dx}\ln(1+x^2)$ at $x=1$." answer="1" hint="Differentiate first, then substitute $x=1$." solution="We have $\qquad \dfrac{d}{dx}\ln(1+x^2)=\dfrac{2x}{1+x^2}$ At $x=1$, $\qquad \dfrac{2}{2}=1$ Hence the answer is $\boxed{1}$." ::: :::question type="MSQ" question="Which of the following are correct?" options=["$\dfrac{d}{dx}\sin(x^3)=3x^2\cos(x^3)$","$\dfrac{d}{dx}e^{x^2}=e^{x^2}$","$\dfrac{d}{dx}\sqrt{1+x}=\dfrac{1}{2\sqrt{1+x}}$","If $s=s(u)$ and $u=u(t)$, then $\dfrac{ds}{dt}=\dfrac{ds}{du}\dfrac{du}{dt}$"] answer="A,C,D" hint="Check chain rule carefully in each case." solution="1. True.

False. The derivative is

$\qquad e^{x^2}\cdot 2x$

True.

True. This is the chain rule in related-rates form.

Hence the correct answer is $\boxed{A,C,D}$." ::: :::question type="SUB" question="Differentiate $y=\cos\big((2x-1)^3\big)$." answer="$-6(2x-1)^2\sin\big((2x-1)^3\big)$" hint="There are two layers inside the cosine." solution="Let $\qquad y=\cos\big((2x-1)^3\big)$ Differentiate outer function first: $\qquad \dfrac{dy}{dx}=-\sin\big((2x-1)^3\big)\cdot \dfrac{d}{dx}\big((2x-1)^3\big)$ Now differentiate the inner cube: $\qquad \dfrac{d}{dx}\big((2x-1)^3\big)=3(2x-1)^2\cdot 2=6(2x-1)^2$ So, $\qquad \dfrac{dy}{dx}=-6(2x-1)^2\sin\big((2x-1)^3\big)$ Hence the derivative is $\boxed{-6(2x-1)^2\sin\big((2x-1)^3\big)}$." ::: ---

Summary

Chapter Summary

Chapter Review Questions

:::question type="MCQ" question="Using first principles, evaluate the derivative of $f(x) = x^2$ at $x=3$." options=["$2x$", "$6$", "$3$", "$9$"] answer="6" hint="Recall the definition $\frac{df}{dx} = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$." solution="The derivative using first principles at $x=3$ is given by:

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:::question type="NAT" question="If $y = e^{\sin(x)}$, determine the value of $\frac{dy}{dx}$ at $x=0$." answer="1" hint="Apply the chain rule $\frac{d}{dx} g(f(x)) = g'(f(x)) \cdot f'(x)$." solution="Given $y = e^{\sin(x)}$, we apply the chain rule.
Let $u = \sin(x)$, then $y = e^u$.
$\frac{dy}{du} = e^u$ and $\frac{du}{dx} = \cos(x)$.
So, $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = e^{\sin(x)} \cdot \cos(x)$.
Now, evaluate at $x=0$:
$\frac{dy}{dx}\Big|_{x=0} = e^{\sin(0)} \cdot \cos(0) = e^0 \cdot 1 = 1 \cdot 1 = 1$."
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:::question type="MCQ" question="For the function $f(x) = \begin{cases} x^2 & x \le 1 \\ 2x-1 & x > 1 \end{cases}$, which statement is true about its differentiability at $x=1$?" options=["The function is continuous but not differentiable at $x=1$.", "The function is differentiable at $x=1$.", "The left-hand derivative exists, but the right-hand derivative does not.", "The function is neither continuous nor differentiable at $x=1$."] answer="The function is differentiable at $x=1$." hint="First, check continuity at $x=1$. Then, evaluate the left-hand and right-hand derivatives at $x=1$." solution="1. Check for Continuity at $x=1$:
$f(1) = 1^2 = 1$.
$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} x^2 = 1^2 = 1$.
$\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (2x-1) = 2(1)-1 = 1$.
Since $f(1) = \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x)$, the function is continuous at $x=1$.

2. Check for Differentiability at $x=1$:
We need to evaluate the left-hand derivative ($f'_L(1)$) and the right-hand derivative ($f'_R(1)$).
For $x < 1$, $f'(x) = \frac{d}{dx}(x^2) = 2x$.
$f'_L(1) = \lim_{x \to 1^-} 2x = 2(1) = 2$.
For $x > 1$, $f'(x) = \frac{d}{dx}(2x-1) = 2$.
$f'_R(1) = \lim_{x \to 1^+} 2 = 2$.
Since $f'_L(1) = f'_R(1) = 2$, the function is differentiable at $x=1$.

Therefore, the function is differentiable at $x=1$."
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What's Next?