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Calculus 1

Differentiation: Definition and Basic Derivative Rules

Finding the Derivatives of Tangent, Cotangent, Secant, And/or Cosecant Functions

Calculus 1

Dec 11, 2025

244

7 pages

Mastering Differential Calculus: Understanding Parametric Equations

user profile picture

Ella Nadine @nadenden

Parametric equations express x and y in terms of a third variable called a parameter. In calculus, finding... Show more

# PARAMETRIC EQUATIONS Given: $y = f(t)$ $x = g(t)$ where y and x are expresses in terms of the same variable known as the parameter. So

Understanding Parametric Equations

Parametric equations are written as x = g(t) and y = f(t), where t is the parameter connecting the two variables. When you need to find derivatives in the regular y-with-respect-to-x form, you'll need a special approach.

To find the first derivative dy/dxdy/dx, follow these steps

  1. Find dy/dt (the derivative of y with respect to the parameter)
  2. Find dx/dt (the derivative of x with respect to the parameter)
  3. Then calculate dy/dx = dy/dtdy/dt ÷ dx/dtdx/dt

For the second derivative d2y/dx2d²y/dx², the process is

  1. Differentiate dy/dx with respect to t
  2. Multiply this result by 1/dx/dtdx/dt

💡 Think of parametric differentiation as using the chain rule in reverse - instead of breaking down a complex derivative, you're building one from simpler parts.

# PARAMETRIC EQUATIONS Given: $y = f(t)$ $x = g(t)$ where y and x are expresses in terms of the same variable known as the parameter. So

Example 1 Finding Parametric Derivatives

Let's work through an example with x = 1 + t² and y = 4t - 3. To find the first derivative, we follow our steps

First, find the individual derivatives with respect to t

  • dx/dt = 2t derivativeof1+t2derivative of 1 + t²
  • dy/dt = 4 derivativeof4t−3derivative of 4t - 3

Now we can find dy/dx by dividing dy/dx = dy/dtdy/dt ÷ dx/dtdx/dt = 4 ÷ 2t = 2/t

For the second derivative, we

  1. Differentiate dy/dx with respect to t d/dt2/t2/t = -2/t²
  2. Multiply by 1/dx/dtdx/dt −2/t2-2/t² × 1/2t1/2t = -1/t³

So our second derivative is d²y/dx² = -1/t³. You can verify this makes sense by thinking about how the curve behaves as t changes.

# PARAMETRIC EQUATIONS Given: $y = f(t)$ $x = g(t)$ where y and x are expresses in terms of the same variable known as the parameter. So

Example 2 Trigonometric Parameters

For the parametric equations x = 2 + sec θ and y = 1 + 2 tan θ, we apply the same principles but with trigonometric functions.

Finding the first derivative

  • dx/dθ = sec θ tan θ derivativeof2+secθderivative of 2 + sec θ
  • dy/dθ = 2 sec² θ derivativeof1+2tanθderivative of 1 + 2 tan θ

Dividing these gives us dy/dx = (2 sec² θ) ÷ (sec θ tan θ) = 2 sec θ/tan θ = 2/sin θ = 2 csc θ

For the second derivative

  1. Differentiate dy/dx with respect to θ d/dθ(2 csc θ) = -2 csc θ cot θ
  2. Multiply by 1/dx/dθdx/dθ −2cscθcotθ-2 csc θ cot θ × 1/secθtanθ1/sec θ tan θ

After algebraic simplification, we get d²y/dx² = -2 cot³ θ.

🔑 When working with trigonometric parametric equations, remember your derivative formulas and be ready to use trigonometric identities to simplify expressions.

# PARAMETRIC EQUATIONS Given: $y = f(t)$ $x = g(t)$ where y and x are expresses in terms of the same variable known as the parameter. So

Example 3 Logarithmic Parameters

Our final example involves logarithms x = ln t² and y = ln ln t. These types of parametric equations often appear in more advanced problems.

For the first derivative

  • First simplify x = ln t² to x = 2 ln t
  • dx/dt = 2/t
  • dy/dt = 1/(t ln t)

The first derivative is dy/dx = 1/(tlnt)1/(t ln t) ÷ 2/t2/t = 1/(2 ln t)

For the second derivative, we

  1. Differentiate dy/dx with respect to t d/dt1/(2lnt)1/(2 ln t) = -1/(2t ln² t)
  2. Multiply by 1/dx/dtdx/dt −1/(2tln2t)-1/(2t ln² t) × t/2t/2 = -1/(4 ln² t)

Our final answer is d²y/dx² = -1/(4 ln² t).

You've now mastered the technique for finding derivatives of parametric equations! This skill will be crucial for analyzing parametric curves and solving related calculus problems.

# PARAMETRIC EQUATIONS Given: $y = f(t)$ $x = g(t)$ where y and x are expresses in terms of the same variable known as the parameter. So
# PARAMETRIC EQUATIONS Given: $y = f(t)$ $x = g(t)$ where y and x are expresses in terms of the same variable known as the parameter. So
# PARAMETRIC EQUATIONS Given: $y = f(t)$ $x = g(t)$ where y and x are expresses in terms of the same variable known as the parameter. So

We thought you’d never ask...

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Is Knowunity really free of charge?

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THE QUIZES AND FLASHCARDS ARE SO USEFUL AND I LOVE THE SCHOOLGPT. IT ALSO IS LITREALLY LIKE CHATGPT BUT SMARTER!! HELPED ME WITH MY MASCARA PROBLEMS TOO!! AS WELL AS MY REAL SUBJECTS ! DUHHH 😍😁😲🤑💗✨🎀😮

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The app is very easy to use and well designed. I have found everything I was looking for so far and have been able to learn a lot from the presentations! I will definitely use the app for a class assignment! And of course it also helps a lot as an inspiration.

Stefan S

iOS user

This app is really great. There are so many study notes and help [...]. My problem subject is French, for example, and the app has so many options for help. Thanks to this app, I have improved my French. I would recommend it to anyone.

Samantha Klich

Android user

Wow, I am really amazed. I just tried the app because I've seen it advertised many times and was absolutely stunned. This app is THE HELP you want for school and above all, it offers so many things, such as workouts and fact sheets, which have been VERY helpful to me personally.

Anna

iOS user

I think it’s very much worth it and you’ll end up using it a lot once you get the hang of it and even after looking at others notes you can still ask your Artificial intelligence buddy the question and ask to simplify it if you still don’t get it!!! In the end I think it’s worth it 😊👍 ⚠️Also DID I MENTION ITS FREEE YOU DON’T HAVE TO PAY FOR ANYTHING AND STILL GET YOUR GRADES IN PERFECTLY❗️❗️⚠️

Thomas R

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Knowunity is the BEST app I’ve used in a minute. This is not an ai review or anything this is genuinely coming from a 7th grade student (I know 2011 im young) but dude this app is a 10/10 i have maintained a 3.8 gpa and have plenty of time for gaming. I love it and my mom is just happy I got good grades

Brad T

Android user

Not only did it help me find the answer but it also showed me alternative ways to solve it. I was horrible in math and science but now I have an a in both subjects. Thanks for the help🤍🤍

David K

iOS user

The app's just great! All I have to do is enter the topic in the search bar and I get the response real fast. I don't have to watch 10 YouTube videos to understand something, so I'm saving my time. Highly recommended!

Sudenaz Ocak

Android user

In school I was really bad at maths but thanks to the app, I am doing better now. I am so grateful that you made the app.

Greenlight Bonnie

Android user

I found this app a couple years ago and it has only gotten better since then. I really love it because it can help with written questions and photo questions. Also, it can find study guides that other people have made as well as flashcard sets and practice tests. The free version is also amazing for students who might not be able to afford it. Would 100% recommend

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Best app if you're in Highschool or Junior high. I have been using this app for 2 school years and it's the best, it's good if you don't have anyone to help you with school work.😋🩷🎀

Marco B

iOS user

THE QUIZES AND FLASHCARDS ARE SO USEFUL AND I LOVE THE SCHOOLGPT. IT ALSO IS LITREALLY LIKE CHATGPT BUT SMARTER!! HELPED ME WITH MY MASCARA PROBLEMS TOO!! AS WELL AS MY REAL SUBJECTS ! DUHHH 😍😁😲🤑💗✨🎀😮

Elisha

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This app is phenomenal down to the correct info and the various topics you can study! I greatly recommend it for people who struggle with procrastination and those who need homework help. It has been perfectly accurate for world 1 history as far as I’ve seen! Geometry too!

Paul T

iOS user

Subjects

Calculus 1

Differentiation: Definition and Basic Derivative Rules

Finding the Derivatives of Tangent, Cotangent, Secant, And/or Cosecant Functions

 

Calculus 1

•

244

•

Dec 11, 2025

•

7 pages

Mastering Differential Calculus: Understanding Parametric Equations

user profile picture

Ella Nadine

@nadenden

Parametric equations express x and y in terms of a third variable called a parameter. In calculus, finding derivatives of parametric equations requires a specific approach that differs from regular differentiation. This guide will show you how to find first... Show more

# PARAMETRIC EQUATIONS Given: $y = f(t)$ $x = g(t)$ where y and x are expresses in terms of the same variable known as the parameter. So

Sign up to see the contentIt's free!

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Understanding Parametric Equations

Parametric equations are written as x = g(t) and y = f(t), where t is the parameter connecting the two variables. When you need to find derivatives in the regular y-with-respect-to-x form, you'll need a special approach.

To find the first derivative dy/dxdy/dx, follow these steps:

  1. Find dy/dt (the derivative of y with respect to the parameter)
  2. Find dx/dt (the derivative of x with respect to the parameter)
  3. Then calculate: dy/dx = dy/dtdy/dt ÷ dx/dtdx/dt

For the second derivative d2y/dx2d²y/dx², the process is:

  1. Differentiate dy/dx with respect to t
  2. Multiply this result by 1/dx/dtdx/dt

💡 Think of parametric differentiation as using the chain rule in reverse - instead of breaking down a complex derivative, you're building one from simpler parts.

# PARAMETRIC EQUATIONS Given: $y = f(t)$ $x = g(t)$ where y and x are expresses in terms of the same variable known as the parameter. So

Sign up to see the contentIt's free!

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Improve your grades

Join milions of students

By signing up you accept Terms of Service and Privacy Policy

Example 1: Finding Parametric Derivatives

Let's work through an example with x = 1 + t² and y = 4t - 3. To find the first derivative, we follow our steps:

First, find the individual derivatives with respect to t:

  • dx/dt = 2t derivativeof1+t2derivative of 1 + t²
  • dy/dt = 4 derivativeof4t−3derivative of 4t - 3

Now we can find dy/dx by dividing: dy/dx = dy/dtdy/dt ÷ dx/dtdx/dt = 4 ÷ 2t = 2/t

For the second derivative, we:

  1. Differentiate dy/dx with respect to t: d/dt2/t2/t = -2/t²
  2. Multiply by 1/dx/dtdx/dt: −2/t2-2/t² × 1/2t1/2t = -1/t³

So our second derivative is d²y/dx² = -1/t³. You can verify this makes sense by thinking about how the curve behaves as t changes.

# PARAMETRIC EQUATIONS Given: $y = f(t)$ $x = g(t)$ where y and x are expresses in terms of the same variable known as the parameter. So

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Example 2: Trigonometric Parameters

For the parametric equations x = 2 + sec θ and y = 1 + 2 tan θ, we apply the same principles but with trigonometric functions.

Finding the first derivative:

  • dx/dθ = sec θ tan θ derivativeof2+secθderivative of 2 + sec θ
  • dy/dθ = 2 sec² θ derivativeof1+2tanθderivative of 1 + 2 tan θ

Dividing these gives us: dy/dx = (2 sec² θ) ÷ (sec θ tan θ) = 2 sec θ/tan θ = 2/sin θ = 2 csc θ

For the second derivative:

  1. Differentiate dy/dx with respect to θ: d/dθ(2 csc θ) = -2 csc θ cot θ
  2. Multiply by 1/dx/dθdx/dθ: −2cscθcotθ-2 csc θ cot θ × 1/secθtanθ1/sec θ tan θ

After algebraic simplification, we get d²y/dx² = -2 cot³ θ.

🔑 When working with trigonometric parametric equations, remember your derivative formulas and be ready to use trigonometric identities to simplify expressions.

# PARAMETRIC EQUATIONS Given: $y = f(t)$ $x = g(t)$ where y and x are expresses in terms of the same variable known as the parameter. So

Sign up to see the contentIt's free!

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Example 3: Logarithmic Parameters

Our final example involves logarithms: x = ln t² and y = ln ln t. These types of parametric equations often appear in more advanced problems.

For the first derivative:

  • First simplify x = ln t² to x = 2 ln t
  • dx/dt = 2/t
  • dy/dt = 1/(t ln t)

The first derivative is: dy/dx = 1/(tlnt)1/(t ln t) ÷ 2/t2/t = 1/(2 ln t)

For the second derivative, we:

  1. Differentiate dy/dx with respect to t: d/dt1/(2lnt)1/(2 ln t) = -1/(2t ln² t)
  2. Multiply by 1/dx/dtdx/dt: −1/(2tln2t)-1/(2t ln² t) × t/2t/2 = -1/(4 ln² t)

Our final answer is d²y/dx² = -1/(4 ln² t).

You've now mastered the technique for finding derivatives of parametric equations! This skill will be crucial for analyzing parametric curves and solving related calculus problems.

# PARAMETRIC EQUATIONS Given: $y = f(t)$ $x = g(t)$ where y and x are expresses in terms of the same variable known as the parameter. So

Sign up to see the contentIt's free!

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# PARAMETRIC EQUATIONS Given: $y = f(t)$ $x = g(t)$ where y and x are expresses in terms of the same variable known as the parameter. So

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# PARAMETRIC EQUATIONS Given: $y = f(t)$ $x = g(t)$ where y and x are expresses in terms of the same variable known as the parameter. So

Sign up to see the contentIt's free!

Access to all documents

Improve your grades

Join milions of students

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We thought you’d never ask...

What is the Knowunity AI companion?

Our AI companion is specifically built for the needs of students. Based on the millions of content pieces we have on the platform we can provide truly meaningful and relevant answers to students. But its not only about answers, the companion is even more about guiding students through their daily learning challenges, with personalised study plans, quizzes or content pieces in the chat and 100% personalisation based on the students skills and developments.

Where can I download the Knowunity app?

You can download the app in the Google Play Store and in the Apple App Store.

Is Knowunity really free of charge?

That's right! Enjoy free access to study content, connect with fellow students, and get instant help – all at your fingertips.

2

Smart Tools NEW

Transform this note into: ✓ 50+ Practice Questions ✓ Interactive Flashcards ✓ Full Mock Exam ✓ Essay Outlines

Mock Exam
Quiz
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Essay

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The app is very easy to use and well designed. I have found everything I was looking for so far and have been able to learn a lot from the presentations! I will definitely use the app for a class assignment! And of course it also helps a lot as an inspiration.

Stefan S

iOS user

This app is really great. There are so many study notes and help [...]. My problem subject is French, for example, and the app has so many options for help. Thanks to this app, I have improved my French. I would recommend it to anyone.

Samantha Klich

Android user

Wow, I am really amazed. I just tried the app because I've seen it advertised many times and was absolutely stunned. This app is THE HELP you want for school and above all, it offers so many things, such as workouts and fact sheets, which have been VERY helpful to me personally.

Anna

iOS user

I think it’s very much worth it and you’ll end up using it a lot once you get the hang of it and even after looking at others notes you can still ask your Artificial intelligence buddy the question and ask to simplify it if you still don’t get it!!! In the end I think it’s worth it 😊👍 ⚠️Also DID I MENTION ITS FREEE YOU DON’T HAVE TO PAY FOR ANYTHING AND STILL GET YOUR GRADES IN PERFECTLY❗️❗️⚠️

Thomas R

iOS user

Knowunity is the BEST app I’ve used in a minute. This is not an ai review or anything this is genuinely coming from a 7th grade student (I know 2011 im young) but dude this app is a 10/10 i have maintained a 3.8 gpa and have plenty of time for gaming. I love it and my mom is just happy I got good grades

Brad T

Android user

Not only did it help me find the answer but it also showed me alternative ways to solve it. I was horrible in math and science but now I have an a in both subjects. Thanks for the help🤍🤍

David K

iOS user

The app's just great! All I have to do is enter the topic in the search bar and I get the response real fast. I don't have to watch 10 YouTube videos to understand something, so I'm saving my time. Highly recommended!

Sudenaz Ocak

Android user

In school I was really bad at maths but thanks to the app, I am doing better now. I am so grateful that you made the app.

Greenlight Bonnie

Android user

I found this app a couple years ago and it has only gotten better since then. I really love it because it can help with written questions and photo questions. Also, it can find study guides that other people have made as well as flashcard sets and practice tests. The free version is also amazing for students who might not be able to afford it. Would 100% recommend

Aubrey

iOS user

Best app if you're in Highschool or Junior high. I have been using this app for 2 school years and it's the best, it's good if you don't have anyone to help you with school work.😋🩷🎀

Marco B

iOS user

THE QUIZES AND FLASHCARDS ARE SO USEFUL AND I LOVE THE SCHOOLGPT. IT ALSO IS LITREALLY LIKE CHATGPT BUT SMARTER!! HELPED ME WITH MY MASCARA PROBLEMS TOO!! AS WELL AS MY REAL SUBJECTS ! DUHHH 😍😁😲🤑💗✨🎀😮

Elisha

iOS user

This app is phenomenal down to the correct info and the various topics you can study! I greatly recommend it for people who struggle with procrastination and those who need homework help. It has been perfectly accurate for world 1 history as far as I’ve seen! Geometry too!

Paul T

iOS user

The app is very easy to use and well designed. I have found everything I was looking for so far and have been able to learn a lot from the presentations! I will definitely use the app for a class assignment! And of course it also helps a lot as an inspiration.

Stefan S

iOS user

This app is really great. There are so many study notes and help [...]. My problem subject is French, for example, and the app has so many options for help. Thanks to this app, I have improved my French. I would recommend it to anyone.

Samantha Klich

Android user

Wow, I am really amazed. I just tried the app because I've seen it advertised many times and was absolutely stunned. This app is THE HELP you want for school and above all, it offers so many things, such as workouts and fact sheets, which have been VERY helpful to me personally.

Anna

iOS user

I think it’s very much worth it and you’ll end up using it a lot once you get the hang of it and even after looking at others notes you can still ask your Artificial intelligence buddy the question and ask to simplify it if you still don’t get it!!! In the end I think it’s worth it 😊👍 ⚠️Also DID I MENTION ITS FREEE YOU DON’T HAVE TO PAY FOR ANYTHING AND STILL GET YOUR GRADES IN PERFECTLY❗️❗️⚠️

Thomas R

iOS user

Knowunity is the BEST app I’ve used in a minute. This is not an ai review or anything this is genuinely coming from a 7th grade student (I know 2011 im young) but dude this app is a 10/10 i have maintained a 3.8 gpa and have plenty of time for gaming. I love it and my mom is just happy I got good grades

Brad T

Android user

Not only did it help me find the answer but it also showed me alternative ways to solve it. I was horrible in math and science but now I have an a in both subjects. Thanks for the help🤍🤍

David K

iOS user

The app's just great! All I have to do is enter the topic in the search bar and I get the response real fast. I don't have to watch 10 YouTube videos to understand something, so I'm saving my time. Highly recommended!

Sudenaz Ocak

Android user

In school I was really bad at maths but thanks to the app, I am doing better now. I am so grateful that you made the app.

Greenlight Bonnie

Android user

I found this app a couple years ago and it has only gotten better since then. I really love it because it can help with written questions and photo questions. Also, it can find study guides that other people have made as well as flashcard sets and practice tests. The free version is also amazing for students who might not be able to afford it. Would 100% recommend

Aubrey

iOS user

Best app if you're in Highschool or Junior high. I have been using this app for 2 school years and it's the best, it's good if you don't have anyone to help you with school work.😋🩷🎀

Marco B

iOS user

THE QUIZES AND FLASHCARDS ARE SO USEFUL AND I LOVE THE SCHOOLGPT. IT ALSO IS LITREALLY LIKE CHATGPT BUT SMARTER!! HELPED ME WITH MY MASCARA PROBLEMS TOO!! AS WELL AS MY REAL SUBJECTS ! DUHHH 😍😁😲🤑💗✨🎀😮

Elisha

iOS user

This app is phenomenal down to the correct info and the various topics you can study! I greatly recommend it for people who struggle with procrastination and those who need homework help. It has been perfectly accurate for world 1 history as far as I’ve seen! Geometry too!

Paul T

iOS user