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Summary of Limits

Michael Green

Michael Green

6 min read

Next Topic - Continuity

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Study Guide Overview

This guide covers calculating derivatives, including estimating derivatives from graphs/tables and finding derivative expressions. It explores basic differentiation, differentiating exponentials, logarithms, and trigonometric functions. It also covers the product, quotient, and chain rules, as well as the inverse function theorem and implicit differentiation.

#Calculating Derivatives: Comprehensive Study Notes

#Table of Contents

  1. Introduction to Derivatives
  2. Estimating Derivatives
  3. Finding the Derivative as an Expression
    • Basic Differentiation
    • Differentiating Exponentials and Logarithms
    • Differentiating Trigonometric Functions
    • Product Rule and Quotient Rule
    • Chain Rule
    • Inverse Function Theorem
    • Implicit Differentiation
  4. Practice Questions
  5. Glossary
  6. Summary and Key Takeaways

#Introduction to Derivatives

Key Concept

Understand the key methods for differentiating functions and selecting the most appropriate method for exam questions.

A derivative of a function represents the rate at which the function's value changes as its input changes. The derivative of f(x)f(x)f(x) is defined as: f′(x)=lim⁡h→0f(x+h)−f(x)h{f}'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}f′(x)=h→0lim​hf(x+h)−f(x)​

Exam Tip

You generally only need to use this definition if specifically asked in an exam.


#Estimating Derivatives

Key Concept

Estimate the derivative of a function at a point using a table or graph.

To estimate a derivative at a point using a table or graph, remember:

  • The derivative at a point equals the slope of the tangent at that point.
  • To approximate the slope of the tangent to the graph of f(x)f(x)f(x) at x=ax = ax=a:
    • Find the slope of line segments joining nearby coordinates that lie on the graph.
Exam Tip

This is particularly useful when you only have a graph or table available.


#Finding the Derivative as an Expression

#Basic Differentiation

Key Concept

Differentiate basic functions and expressions.

  1. Powers of xxx:
    • If f(x)=xnf(x) = x^nf(x)=xn, then f′(x)=nxn−1f'(x) = nx^{n-1}f′(x)=nxn−1
  2. Sums and Differences:
    • The derivative of a sum (or difference) of terms is the sum (or difference) of the derivatives of the terms.
  3. Constant Multiples:
    • If f(x)=axnf(x) = ax^nf(x)=axn, then f′(x)=anxn−1f'(x) = an x^{n-1}f′(x)=anxn−1
  4. Special Cases:
    • If f(x)=axf(x) = axf(x)=ax, then f′(x)=af'(x) = af′(x)=a
    • If g(x)=ag(x) = ag(x)=a, then g′(x)=0g'(x) = 0g′(x)=0
Common Mistake

Remember to expand brackets or simplify expressions before differentiating. For example, rewrite x\sqrt{x}x​ as x1/2x^{1/2}x1/2.

#Differentiating Exponentials and Logarithms

Key Concept

Differentiate exponential and logarithmic functions.

f(x)f(x)f(x)f′(x)f'(x)f′(x)
ekxe^{kx}ekxkekxke^{kx}kekx
akxa^{kx}akxakxkln⁡aa^{kx} k \ln aakxklna
ln⁡(kx)\ln(kx)ln(kx)1x\frac{1}{x}x1​

#Differentiating Trigonometric Functions

Key Concept

Differentiate trigonometric and inverse trigonometric functions.

f(x)f(x)f(x)f′(x)f'(x)f′(x)
sin⁡(kx)\sin(kx)sin(kx)kcos⁡(kx)k \cos(kx)kcos(kx)
cos⁡(kx)\cos(kx)cos(kx)−ksin⁡(kx)-k \sin(kx)−ksin(kx)
tan⁡(kx)\tan(kx)tan(kx)ksec⁡2(kx)k \sec^2(kx)ksec2(kx)
csc⁡(kx)\csc(kx)csc(kx)−kcot⁡(kx)csc⁡(kx)-k \cot(kx) \csc(kx)−kcot(kx)csc(kx)
sec⁡(kx)\sec(kx)sec(kx)ktan⁡(kx)sec⁡(kx)k \tan(kx) \sec(kx)ktan(kx)sec(kx)
cot⁡(kx)\cot(kx)cot(kx)−kcsc⁡2(kx)-k \csc^2(kx)−kcsc2(kx)
arcsin⁡(x)\arcsin(x)arcsin(x)11−x2,−1<x<1\frac{1}{\sqrt{1 - x^2}}, -1 < x < 11−x2​1​,−1<x<1
arccos⁡(x)\arccos(x)arccos(x)−11−x2,−1<x<1-\frac{1}{\sqrt{1 - x^2}}, -1 < x < 1−1−x2​1​,−1<x<1
arctan⁡(x)\arctan(x)arctan(x)11+x2\frac{1}{1 + x^2}1+x21​
Key Concept

The most important results to remember are for sin⁡\sinsin and cos⁡\coscos. The results for tan⁡\tantan, csc⁡\csccsc, sec⁡\secsec, and cot⁡\cotcot can all be derived using the quotient rule and trigonometric identities.

#Product Rule and Quotient Rule

Key Concept

Apply the product rule and quotient rule to differentiate functions.

  1. Product Rule:

    • If f(x)=g(x)⋅h(x)f(x) = g(x) \cdot h(x)f(x)=g(x)⋅h(x), then: f′(x)=g′(x)⋅h(x)+g(x)⋅h′(x)f'(x) = g'(x) \cdot h(x) + g(x) \cdot h'(x)f′(x)=g′(x)⋅h(x)+g(x)⋅h′(x)
    • Alternatively, if y=u⋅vy = u \cdot vy=u⋅v, then: dydx=dudx⋅v+u⋅dvdx\frac{dy}{dx} = \frac{du}{dx} \cdot v + u \cdot \frac{dv}{dx}dxdy​=dxdu​⋅v+u⋅dxdv​ y′=u′v+uv′y' = u'v + uv'y′=u′v+uv′
  2. Quotient Rule:

    • If f(x)=g(x)h(x)f(x) = \frac{g(x)}{h(x)}f(x)=h(x)g(x)​, then: f′(x)=g′(x)⋅h(x)−g(x)⋅h′(x)(h(x))2f'(x) = \frac{g'(x) \cdot h(x) - g(x) \cdot h'(x)}{(h(x))^2}f′(x)=(h(x))2g′(x)⋅h(x)−g(x)⋅h′(x)​
    • Alternatively, if y=uvy = \frac{u}{v}y=vu​, then: dydx=dudx⋅v−u⋅dvdxv2\frac{dy}{dx} = \frac{\frac{du}{dx} \cdot v - u \cdot \frac{dv}{dx}}{v^2}dxdy​=v2dxdu​⋅v−u⋅dxdv​​ y′=u′v−uv′v2y' = \frac{u'v - uv'}{v^2}y′=v2u′v−uv′​

#Chain Rule

Key Concept

Use the chain rule for differentiating composite functions.

  1. Chain Rule:
    • If y=f(u)y = f(u)y=f(u) and u=g(x)u = g(x)u=g(x), then: dydx=dydu⋅dudx\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}dxdy​=dudy​⋅dxdu​
    • In function notation, if h(x)=f(g(x))h(x) = f(g(x))h(x)=f(g(x)), then: h′(x)=f′(g(x))⋅g′(x)h'(x) = f'(g(x)) \cdot g'(x)h′(x)=f′(g(x))⋅g′(x)

#Inverse Function Theorem

Key Concept

Apply the inverse function theorem to find the derivative of an inverse function.

  1. Inverse Function Theorem:
    • For a function fff, the derivative of its inverse is given by: (f−1)′(a)=1f′(f−1(a))\left( f^{-1} \right)'(a) = \frac{1}{f'(f^{-1}(a))}(f−1)′(a)=f′(f−1(a))1​
    • Alternatively, if g(a)=f−1(a)g(a) = f^{-1}(a)g(a)=f−1(a): g′(a)=1f′(g(a))g'(a) = \frac{1}{f'(g(a))}g′(a)=f′(g(a))1​
    • If y=f−1(x)y = f^{-1}(x)y=f−1(x) so that x=f(y)x = f(y)x=f(y), then: dydx=1(dxdy)\frac{dy}{dx} = \frac{1}{\left( \frac{dx}{dy} \right)}dxdy​=(dydx​)1​

#Implicit Differentiation

Key Concept

Differentiate functions written implicitly.

  1. Implicit Differentiation:
    • For equations like 3x2−7xy2=33x^2 - 7xy^2 = 33x2−7xy2=3 or x2+y2=25x^2 + y^2 = 25x2+y2=25:
      • Every term in the equation is differentiated.
      • For terms involving yyy, apply the chain rule: ddxf(y)=f′(y)⋅y′=f′(y)⋅dydx\frac{d}{dx} f(y) = f'(y) \cdot y' = f'(y) \cdot \frac{dy}{dx}dxd​f(y)=f′(y)⋅y′=f′(y)⋅dxdy​
Exam Tip

Differentiate the function in terms of yyy with respect to yyy, then multiply by dydx\frac{dy}{dx}dxdy​.


#Practice Questions

Practice Question
  1. Differentiate f(x)=3x4−5x2+2x−7f(x) = 3x^4 - 5x^2 + 2x - 7f(x)=3x4−5x2+2x−7.
Practice Question
  1. Find the derivative of g(x)=e2xg(x) = e^{2x}g(x)=e2x.
Practice Question
  1. Use the product rule to differentiate h(x)=(3x2)(sin⁡x)h(x) = (3x^2)(\sin x)h(x)=(3x2)(sinx).
Practice Question
  1. Apply the chain rule to differentiate y=sin⁡(3x2)y = \sin(3x^2)y=sin(3x2).
Practice Question
  1. Differentiate implicitly: x2+y2=1x^2 + y^2 = 1x2+y2=1.

#Glossary

  • Derivative: The rate of change of a function with respect to its variable.
  • Tangent: A line that touches a curve at a point without crossing over.
  • Product Rule: A rule for differentiating products of two functions.
  • Quotient Rule: A rule for differentiating quotients of two functions.
  • Chain Rule: A rule for differentiating composite functions.
  • Implicit Differentiation: A method for differentiating functions that are not explicitly solved for one variable.

#Summary and Key Takeaways

  • Understand the various methods for differentiating functions.
  • Use the definition of the derivative only if explicitly asked.
  • Differentiate basic functions, exponentials, logarithms, and trigonometric functions using their respective rules.
  • Apply the product rule, quotient rule, and chain rule where appropriate.
  • Use the inverse function theorem and implicit differentiation for more complex functions.
  • Practice these techniques regularly to build confidence and proficiency.
Exam Tip

Exam questions often combine multiple differentiation techniques. Break the problem into smaller parts and work methodically.

Exam Tip

Be comfortable switching between different notational forms of the derivative as it helps in remembering and applying rules correctly.


By mastering these concepts, students will be well-prepared to tackle a variety of differentiation problems in their exams.

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Question 1 of 11

What does the derivative of a function represent? 🤔

The area under the curve

The slope of the secant line

The rate at which the function's value changes

The y-intercept of the function