Calculus formulas for all your calculus problems
What is calculus?
It is a branch of mathematics in which we analyze how the rate of the dependent variable changes with respect to the independent variable. In simple words, calculus is the study of the rate of changes. It has been divided into two branches namely, differential calculus and integral calculus. Differential calculus deals with instantaneous rates of change whereas integral calculus covers accumulation of quantities, and areas under or between curves etc.
As per the US national school curriculum, calculus as a topic is introduced in High school. But depending on the school as well as the student’s interest one can start learning about calculus before starting high school through honors courses. Also high school students interested in learning calculus in detail and at an advanced level can take AP calculus courses as it covers topics from college level.
What is calculus 1 & 2 and how are they different?
Calculus 1 covers problems related to differential calculus. So one learns all differentiation rules and how to differentiate functions and composite functions as well as solve using approximations.
Calculus 2 builds techniques learnt in Calculus 1 and relates to integration problems. Students in calculus 2 course learn how to integrate various types of functions, how to find arc length of a curve and surface area of revolution etc.
This blog briefly walks through the calculus rules, some example problems with step-by-step explanations. So let’s get started:
Calculus Formulas
- Function Operation:
Addition | (f+g)(x) = f(x) + g(x) |
Subtraction | (f-g)(x) = f(x) – g(x) |
Multiplication | (f·g)(x) = f(x) · g(x) |
Division | (f/g)(x) = f(x) / g(x) |
- Composition of Function:
(f °g)(x) = f(g(x)) | (g ᵒ f)(x) = g(f(x)) |
- Inverse of Function:
f-1( f(x) ) = x | f(f-1 (x) ) = x |
- Function Transformation:
Vertical Shift | f (x) + b; moves b units upward
f x- b; moves b units downward |
Horizontal Shift | f x + b; shift b units left
f x- b; shift b units right |
Reflection | –f (x); over x-axis
f (-x); over y-axis |
Horizontal Dilation | fxA; if A>1, Away from y-axis by a factor of A.
if 0<A<1, contract by factor 1A. |
Vertical Dilation | Bfx; if B>1, Away from x-axis by a factor of B.
if 0<B<1, contract toward x-axis by a factor 1B. |
- Degree to Radian:
t=πx180° | x=180°t |
Where, x is in degree, and t is in radian.
- Coterminal Angles:
θ±360° | θ±2π |
- Trigonometric Ratio of Reference Angles:
Degree | 0° | 30° | 45° | 60° | 90° |
Radians | 0 | 6 | 4 | 3 | 2 |
sin | 0 | 12 | 1√2 | √32 | 1 |
cos | 1 | √32 | 1√2 | 12 | 0 |
tan | 0 | 1√3 | 1 | √3 | ∞ |
cot | ∞ | √3 | 1 | 1√3 | 0 |
csc | ∞ | 2 | √2 | 1 | |
sec | 1 | 2√3 | √2 | 2 | ∞ |
- Graph sine Function:
y=A sinBx-C+D | |
Period is 2π|B|,
|A| is amplitude. |
C is phase shift.
D is vertical shift. |
CB>0: Translated Left | CB<0: Translated Right |
D>0: Translated Up | D<0: Translated Down |
- Graph cosine Function:
y=A cosBx-C+D | |
Period is 2π|B|,
|A| is amplitude. |
C is phase shift.
D is vertical shift. |
CB>0: Translated Left | CB<0: Translated Right |
D>0: Translated Up | D<0: Translated Down |
- Properties of Logs:
- Properties of Exponents:
- Properties of Limit:
One-sided limits
x slightly greater than a then f(x) is close to l x slightly less than a then f(x) is close to l |
fx=l
fx=l |
Sum of the limits | fx+gx=f(x)+ g(x) |
Difference of the limits | fx-gx=f(x)- g(x) |
Multiple of the limits | c∙fx =c∙fx |
Product of the limits | fx∙gx=f(x) g(x) |
The quotient of the limits | f(x)g(x)=f(x) g(x)
If gx≠0 |
Condition for continuity | The function is defined.
fx=l fx=l and fx=l |
- Limits Involving Infinity:
Vertical asymptote at x=a, if one of these conditions is true. | fx=∞
fx=-∞ fx=-∞ fx=∞ fx=∞ fx=-∞ |
Horizontal asymptote at y=L | fx=L
fx=L |
- Intermediate Value Theorem:
For any value L in between f (a) and f (b), there will be a value c in [a,b] for which f(c) = L.
- Rate of Change:
- Derivatives Rules:
- Inverse Function Rule:
- Derivative of Trigonometric Function:
- Derivative of Inverse Trigonometric Function:
- Derivative Strategies:
How to use the calculus formulas to solve calculus problems?
Having access to a calculus formula sheet is magical but one needs practice with a number of questions to feel comfortable using the formula sheet. Therefore, in order to set you up for long term success, we covered the calculus formulas above and now we will show you some calculus problems and their solutions step by step.
Example 1: Let f(x)=x^{3}e^{7x} on the domain [-1, 1].
(a) Find the x-value(s) of the critical point(s) of f and classify each as a local maximum, a local minimum, or neither, using a number line or similar method to justify your answers.
Explanation: To find the critical point(s), we need to set the first derivative equal to zero, and solve for x.
Step 1: Find f'(x).
Given f(x)=x^{3}e^{7x}, we will use the product rule to compute its derivative.
Step 2: Find the critical point(s).
x^{2}e^{7x}(7x+3)=0
Now, using the Zero Factor Principle:
x^{2}e^{7x }= 0
x = 0, x = 0
And, 7x + 3 = 0
x=-3/7
Let us check these critical points with the domain.
Given domain of the function is [-1,1]. So, let us select the critical point that lie in this domain.
We can see, both points are on the given domain.
Step 3: Identify maximum, minimum.
Final answer: From the above table, we can conclude that, at both the critical point, the function will have a local minimum.
(b) Find the x-value(s) of all global extrema of f and classify each as a global maximum or global minimum, providing justification for your answers.
Explanation: To find the global extrema, we need to check the value of the function at the critical point, and at the endpoint of the interval.
We have critical points at x=0, x=-3/7. And, the given interval is [-1,1].
Step 1: Find the value of the function.
Final answer: From the above table, we can see the function will have a maximum value at x=1, so this point would be the global maximum. And, the function attains a minimum value at x=-3/7, so at this point, the function will have a global minimum.
Example 2: The dimension of a metal sheet is 120 cm by 90 cm. Find the height of the box, that will give the maximum possible volume of the box.
Explanation: We need to set up an equation for volume and use the optimization technique.
Formula used: The volume of the box would be V=lbh
Given: The given height is h cm. The length after the cut would be 120-h and, the width would be 90-2h .
Calculation:
Step 1: Calculate the volume.
Step 2: Taking the derivative on both sides.
Step 3: Finding critical points.
Step 4: Check for values.
The largest value of h would be 45 cm. We can see from the 90 cm side. So, the suitable value is h=16.97cm
Final answer: So, the h=16.97cm would give us the maximum volume of the box
Example 3: What would be the maximum area of a rectangular garden covered by fencing of 150 ft.
Explanation: We need to set up an equation for the area of a rectangle. And using the perimeter, we will find the critical point.
Formula used: The area of the rectangular garden would be A=xy
Given: The given fencing length is 150 ft, which is the garden’s perimeter.
Calculation:
Step 1: Find the relation between the sides.
150 = 2(x + y)
75= x + y
x= 75 – y
Step 2: Taking the derivative on both sides of the area.
Step 3: Finding critical points.
Step 4: Find the maximum area.
Final answer: The largest value of the area would be 1406.25 ft2
Example 4: Determine the derivative of function y = ln( 3x )with respect to x.
Explanation: To determine the derivative of function y = ln( 3x ), we need to apply the chain rule, as this is a composite function.
Formula used: If y = f (g ( x )), the derivative would be dy/dx = f'(g ( x ))g'( x )
Calculation:
Step 1: Apply the chain rule.
Final answer: The derivative of function y = ln(3x) would be dy/dx=( 1/x )
Example 5: Determine the instantaneous acceleration of the velocity function
v(t)=3t^{2}+5 ft/s, at t = 2 sec.
Explanation: To find the instantaneous acceleration, we need to find the derivative of velocity function.
Formula used: Using the power rule d(x^{n})/dx=nx^{n-1} , and we know the derivative of the constant is zero.
Calculation:
Step 1: Calculate the derivative.
Step 2: Find acceleration at t = 2 sec.
a (t) = 6t
a (2) = 6(2) = 12 ft / s^{2}
Final answer: The instantaneous acceleration at t = 2 sec would be 12 ft / s^{2}
Summary:
Hopefully by now you have a good hang of all calculus formulas and how to solve different types of calculus questions but in case you have any specific advanced level calculus problem which you still aren’t able to solve independently then don’t worry, we have got you covered.
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