Continuity and Differentiabilty: Chapter 5 Links
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5.1 | 5.2 | 5.3 | 5.4 | 5.5 | 5.7 | Misc
Parametric Differentiation 5.6 deals with functions in which both x and y are expressed in terms of a third variable, called a parameter. If x=f(t) and y=g(t), where t is a parameter, then the derivative \frac{dy}{dx} can be found using the relation
\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}, \qquad \frac{dx}{dt}\neq 0
In this exercise, we will learn how to differentiate functions given in parametric form and apply the concept to solve various NCERT questions.
Key Concepts
1. Parametric Equations
When both x and y are expressed in terms of a third variable t, called a parameter, the equations are said to be in parametric form.
x=f(t), \qquad y=g(t)
Here t is the parameter and both x and y depend on it.
Example: x=a\cos t,\quad y=a\sin t
2. Derivative of Parametric Functions
If
x=f(t), \qquad y=g(t)
then the derivative of y with respect to x is given by
\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}, \qquad \frac{dx}{dt}\neq0
This formula is obtained using the Chain Rule of Differentiation.
3. Second Derivative in Parametric Form
If \frac{dy}{dx} is expressed as a function of t, then the second derivative is given by
\displaystyle \frac{d^2y}{dx^2}=\frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}
This result is useful for finding higher-order derivatives of curves represented in parametric form.
4. Steps for Parametric Differentiation
- Differentiate y with respect to the parameter t to find \frac{dy}{dt}.
- Differentiate x with respect to t to find \frac{dx}{dt}.
- Use \displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}} to obtain the required derivative.
- For second derivatives, differentiate \frac{dy}{dx} with respect to t and divide by \frac{dx}{dt}.
- To find \frac{d^2y}{dx^2}, differentiate \frac{dy}{dx} with respect to the parameter t and then divide by \frac{dx}{dt}. This follows from the Chain Rule.
Using the Chain Rule,
\displaystyle \frac{d}{dt}\left(\frac{dy}{dx}\right)=\frac{d}{dx}\left(\frac{dy}{dx}\right)\cdot\frac{dx}{dt}=\frac{d^2y}{dx^2}\cdot\frac{dx}{dt}
Therefore,
\displaystyle \frac{d^2y}{dx^2}=\frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}
Tip: While solving questions on Parametric Differentiation, it is often helpful to first find \frac{dy}{dt} and then \frac{dx}{dt}. Thinking of them as the numerator and denominator respectively makes it easier to remember the formula
\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}
and helps avoid interchanging the numerator and denominator.
Let us now solve all the NCERT questions step by step in this exercise 5.6 of Continuity and Differentiability.
Question 1: Parametric Differentiation 5.6
1. Find \displaystyle \frac{dy}{dx} if x=2at^2,\; y=at^4.
Solution
Given,
x=2at^2,\qquad y=at^4
Differentiating y w.r.t. t, we get
\displaystyle \frac{dy}{dt}=4at^3 \qquad ...(1)
Differentiating x w.r.t. t, we get
\displaystyle \frac{dx}{dt}=4at \qquad ...(2)
Using the formula
\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}
Dividing (1) by (2), we get
\displaystyle \frac{dy}{dx}=\frac{4at^3}{4at}
\displaystyle =t^2
Hence,
\boxed{\displaystyle \frac{dy}{dx}=t^2}
Question 2: Parametric Differentiation
2. Find \displaystyle \frac{dy}{dx} if x=a\cos\theta,\; y=b\cos\theta.
Solution
Given,
x=a\cos\theta,\qquad y=b\cos\theta
Differentiating y w.r.t. \theta, we get
\displaystyle \frac{dy}{d\theta}=-b\sin\theta \qquad ...(1)
Differentiating x w.r.t. \theta, we get
\displaystyle \frac{dx}{d\theta}=-a\sin\theta \qquad ...(2)
Using the formula
\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}
Dividing (1) by (2), we get
\displaystyle \frac{dy}{dx}=\frac{-b\sin\theta}{-a\sin\theta}
\displaystyle =\frac{b}{a}
Hence,
\boxed{\displaystyle \frac{dy}{dx}=\frac{b}{a}}
You may already be following Maths Better for NCERT Solutions for the topics like
Likewise this Exercise 5.6 of Continuity and Differentiability for Class 12 Maths is designed to strengthen your concepts and improve step-by-step problem-solving skills. Now, let’s move on to the next question.
Question 3: Continuity and Differentiability 5.6
3. Find \displaystyle \frac{dy}{dx} if x=\sin t,\; y=\cos 2t.
Solution
Given,
x=\sin t,\qquad y=\cos 2t
Differentiating y w.r.t. t, we get
\displaystyle \frac{dy}{dt}=-2\sin 2t \qquad ...(1)
Differentiating x w.r.t. t, we get
\displaystyle \frac{dx}{dt}=\cos t \qquad ...(2)
Using the formula
\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}
Dividing (1) by (2), we get
\displaystyle \frac{dy}{dx}=\frac{-2\sin 2t}{\cos t}
Using \sin 2t=2\sin t\cos t,
\displaystyle \frac{dy}{dx}=\frac{-2(2\sin t\cos t)}{\cos t}
\displaystyle =-4\sin t
Hence,
\boxed{\displaystyle \frac{dy}{dx}=-4\sin t}
Question 4: Parametric Differentiation 5.6
4. Find \displaystyle \frac{dy}{dx} if x=4t,\; y=\frac{4}{t}.
Solution
Given,
x=4t,\qquad y=\frac{4}{t}=4t^{-1}
Differentiating y w.r.t. t, we get
\displaystyle \frac{dy}{dt}=4(-1)t^{-2}=-\frac{4}{t^2} \qquad ...(1)
Differentiating x w.r.t. t, we get
\displaystyle \frac{dx}{dt}=4 \qquad ...(2)
Using the formula
\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}
Dividing (1) by (2), we get
\displaystyle \frac{dy}{dx}=\frac{-\frac{4}{t^2}}{4}
\displaystyle =-\frac{1}{t^2}
Hence,
\boxed{\displaystyle \frac{dy}{dx}=-\frac{1}{t^2}}
Question 5: Parametric Differentiation 5.6
5. Find \displaystyle \frac{dy}{dx} if x=\cos\theta-\cos2\theta,\; y=\sin\theta-\sin2\theta.
Solution
Given,
x=\cos\theta-\cos2\theta,\qquad y=\sin\theta-\sin2\theta
Differentiating y w.r.t. \theta, we get
\displaystyle \frac{dy}{d\theta}=\cos\theta-2\cos2\theta \qquad ...(1)
Differentiating x w.r.t. \theta, we get
\displaystyle \frac{dx}{d\theta}=-\sin\theta+2\sin2\theta \qquad ...(2)
Using the formula
\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}
Dividing (1) by (2), we get
\displaystyle \frac{dy}{dx}=\frac{\cos\theta-2\cos2\theta}{-\sin\theta+2\sin2\theta}
Hence,
\boxed{\displaystyle \frac{dy}{dx}=\frac{\cos\theta-2\cos2\theta}{2\sin2\theta-\sin\theta}}
Important: In parametric form, both x and y are expressed in terms of a third variable called a parameter. To differentiate such functions, we first find \frac{dy}{dt} and \frac{dx}{dt}, and then use these derivatives to obtain \frac{dy}{dx}.
Question 6: Continuity and Differentiability 5.6
6. Find \displaystyle \frac{dy}{dx} if x=a(\theta-\sin\theta),\; y=a(1+\cos\theta).
Solution
Given,
x=a(\theta-\sin\theta),\qquad y=a(1+\cos\theta)
Differentiating y w.r.t. \theta, we get
\displaystyle \frac{dy}{d\theta}=a(0-\sin\theta)=-a\sin\theta \qquad ...(1)
Differentiating x w.r.t. \theta, we get
\displaystyle \frac{dx}{d\theta}=a(1-\cos\theta) \qquad ...(2)
Using the formula
\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}
Dividing (1) by (2), we get
\displaystyle \frac{dy}{dx}=\frac{-a\sin\theta}{a(1-\cos\theta)}
\displaystyle =-\frac{\sin\theta}{1-\cos\theta}
Multiplying numerator and denominator by (1+\cos\theta),
We get, \displaystyle \frac{dy}{dx}=-\frac{\sin\theta(1+\cos\theta)}{1-\cos^2\theta}
\displaystyle =-\frac{\sin\theta(1+\cos\theta)}{\sin^2\theta}
\displaystyle =-\frac{1+\cos\theta}{\sin\theta}
Using the identities 1+\cos\theta=2\cos^2\frac{\theta}{2} and \sin\theta=2\sin\frac{\theta}{2}\cos\frac{\theta}{2},
We get, \displaystyle \frac{dy}{dx} =-\frac{2\cos^2\frac{\theta}{2}}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}
\displaystyle =-\frac{\cos\frac{\theta}{2}}{\sin\frac{\theta}{2}}
\displaystyle =-\cot\frac{\theta}{2}
Hence,
\boxed{\displaystyle \frac{dy}{dx}=-\cot\frac{\theta}{2}}
Question 7: Parametric Differentiation 5.6
7. Find \displaystyle \frac{dy}{dx} if \displaystyle x=\frac{\sin^3 t}{\sqrt{\cos2t}},\; y=\frac{\cos^3 t}{\sqrt{\cos2t}}.
Solution
Given,
\displaystyle x=\frac{\sin^3 t}{\sqrt{\cos2t}},\qquad y=\frac{\cos^3 t}{\sqrt{\cos2t}}
Differentiating y w.r.t. t using the Quotient Rule, we get
\displaystyle \frac{dy}{dt}=\frac{\sqrt{\cos2t}\,\frac{d}{dt}(\cos^3 t)-\cos^3 t\,\frac{d}{dt}(\sqrt{\cos2t})}{\cos2t}
\displaystyle =\frac{\sqrt{\cos2t}(-3\cos^2 t\sin t)-\cos^3 t\left(-\frac{\sin2t}{\sqrt{\cos2t}}\right)}{\cos2t}
Simplifying by taking LCM in the numerator
\displaystyle =\frac{-3\cos^2 t\sin t(\cos2t)+\cos^3 t\sin2t}{(\cos2t)^{3/2}}
\displaystyle =\frac{\cos^2 t\sin t\big(-3\cos2t+2\cos^2 t\big)}{(\cos2t)^{3/2}}
Using \cos2t=2\cos^2 t-1,
We get, \displaystyle -3\cos2t+2\cos^2 t=-3(2\cos^2 t-1)+2\cos^2 t
\displaystyle =3-4\cos^2 t
\displaystyle =-(4\cos^2 t-3)
Also using, \cos3t=4\cos^3 t-3\cos t,
We obtain, \displaystyle 4\cos^2 t-3=\frac{\cos3t}{\cos t}
Therefore, after putting this value, we get
\displaystyle \frac{dy}{dt}=-\frac{\cos^2 t\sin t}{(\cos2t)^{3/2}}\cdot\frac{\cos3t}{\cos t}
\displaystyle =-\frac{\sin t\cos t\,\cos3t}{(\cos2t)^{3/2}} \qquad ...(1)
Differentiating x w.r.t. t using the Quotient Rule, we get
\displaystyle \frac{dx}{dt}=\frac{\sqrt{\cos2t}\,\frac{d}{dt}(\sin^3 t)-\sin^3 t\,\frac{d}{dt}(\sqrt{\cos2t})}{\cos2t}
\displaystyle =\frac{\sqrt{\cos2t}(3\sin^2 t\cos t)-\sin^3 t\left(-\frac{\sin2t}{\sqrt{\cos2t}}\right)}{\cos2t}
Simplifying by taking LCM in the numerator as above
\displaystyle =\frac{3\sin^2 t\cos t(\cos2t)+\sin^3 t\sin2t}{(\cos2t)^{3/2}}
\displaystyle =\frac{\sin^2 t\cos t\big(3\cos2t+2\sin^2 t\big)}{(\cos2t)^{3/2}}
Now using \cos2t=1-2\sin^2 t,
\displaystyle 3\cos2t+2\sin^2 t=3(1-2\sin^2 t)+2\sin^2 t
\displaystyle =3-4\sin^2 t
Also using \sin3t=3\sin t-4\sin^3 t,
We obtain, \displaystyle 3-4\sin^2 t=\frac{\sin3t}{\sin t}
Therefore, we get
\displaystyle \frac{dx}{dt}=\frac{\sin^2 t\cos t}{(\cos2t)^{3/2}}\cdot\frac{\sin3t}{\sin t}
\displaystyle =\frac{\sin t\cos t\,\sin3t}{(\cos2t)^{3/2}} \qquad ...(2)
Using the formula
\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}
Dividing (1) by (2), we get
\displaystyle \frac{dy}{dx}=\frac{-\frac{\sin t\cos t\,\cos3t}{(\cos2t)^{3/2}}}{\frac{\sin t\cos t\,\sin3t}{(\cos2t)^{3/2}}}
\displaystyle =-\frac{\cos3t}{\sin3t}=-\cot3t
Hence,
\boxed{\displaystyle \frac{dy}{dx}=-\cot3t}
NCERT Class 12 Maths includes several important exercises across both Part 1 and Part 2, and I’ll cover them one by one with clear explanations and step-by-step solutions. Many of these questions are also available in video format on my YouTube channel, @Mathsbetter, to help you understand the concepts more visually.
Now, let’s proceed to the next question.
Question 8: Parametric Functions
8. Find \displaystyle \frac{dy}{dx} if x=a\left(\cos t+\log\tan\frac{t}{2}\right),\; y=a\sin t.
Solution
Given,
x=a\left(\cos t+\log\tan\frac{t}{2}\right),\qquad y=a\sin t
Differentiating y w.r.t. t, we get
\displaystyle \frac{dy}{dt}=a\cos t \qquad ...(1)
Differentiating x w.r.t. t, we get
\displaystyle \frac{dx}{dt}=a\left(-\sin t+\frac{d}{dt}\left(\log\tan\frac{t}{2}\right)\right)
Now,
\displaystyle \frac{d}{dt}\left(\log\tan\frac{t}{2}\right)=\frac{1}{\sin t}
Therefore,
\displaystyle \frac{dx}{dt}=a\left(-\sin t+\frac{1}{\sin t}\right)
\displaystyle =a\left(\frac{1-\sin^2 t}{\sin t}\right)
i.e. \displaystyle \frac{dx}{dt}=a\frac{\cos^2 t}{\sin t} \qquad ...(2)
Using the formula
\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}
Dividing (1) by (2), we get
\displaystyle \frac{dy}{dx}=\frac{a\cos t}{a\frac{\cos^2 t}{\sin t}}
\displaystyle =\frac{\sin t}{\cos t}=\tan t
Hence,
\boxed{\displaystyle \frac{dy}{dx}=\tan t}
In many situations, the relationship between x and y is more conveniently described by expressing both variables in terms of a parameter. Parametric differentiation provides a systematic method for finding \frac{dy}{dx} and studying the behaviour of such curves.
Question 9: Parametric Differentiation 5.6
9. Find \displaystyle \frac{dy}{dx} if x=a\sec\theta,\; y=b\tan\theta.
Solution
Given,
x=a\sec\theta,\qquad y=b\tan\theta
Differentiating y w.r.t.\theta, we get
\displaystyle \frac{dy}{d\theta}=b\sec^2\theta \qquad ...(1)
Differentiating x w.r.t. \theta, we get
\displaystyle \frac{dx}{d\theta}=a\sec\theta\tan\theta \qquad ...(2)
Using the formula
\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}
Dividing (1) by (2), we get
\displaystyle \frac{dy}{dx}=\frac{b\sec^2\theta}{a\sec\theta\tan\theta}
\displaystyle =\frac{b}{a}\cdot\frac{\sec\theta}{\tan\theta}
or
\displaystyle =\frac{b}{a}\cdot\frac{\frac{1}{\cos\theta}}{\frac{\sin\theta}{\cos\theta}}
\displaystyle =\frac{b}{a\sin\theta}
or
\displaystyle =\frac{b}{a}\cosec\theta
Hence,
\boxed{\displaystyle \frac{dy}{dx}=\frac{b}{a}\cosec\theta}
Question 10: Parametric Differentiation 5.6
10. Find \displaystyle \frac{dy}{dx} if x=a(\cos\theta+\theta\sin\theta),\; y=a(\sin\theta-\theta\cos\theta).
Solution
Given,
x=a(\cos\theta+\theta\sin\theta),\qquad y=a(\sin\theta-\theta\cos\theta)
Differentiating y w.r.t. \theta, we get
\displaystyle \frac{dy}{d\theta}=a\left(\cos\theta-\frac{d}{d\theta}(\theta\cos\theta)\right)
Using the Product Rule,
\displaystyle \frac{d}{d\theta}(\theta\cos\theta)=\cos\theta-\theta\sin\theta
Therefore,
\displaystyle \frac{dy}{d\theta}=a\big(\cos\theta-\cos\theta+\theta\sin\theta\big)
\displaystyle =a\theta\sin\theta \qquad ...(1)
Differentiating x w.r.t. \theta, we get
\displaystyle \frac{dx}{d\theta}=a\left(-\sin\theta+\frac{d}{d\theta}(\theta\sin\theta)\right)
Using the Product Rule,
\displaystyle \frac{d}{d\theta}(\theta\sin\theta)=\sin\theta+\theta\cos\theta
Therefore,
\displaystyle \frac{dx}{d\theta}=a\big(-\sin\theta+\sin\theta+\theta\cos\theta\big)
\displaystyle =a\theta\cos\theta \qquad ...(2)
Using the formula
\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}
Dividing (1) by (2), we get
\displaystyle \frac{dy}{dx}=\frac{a\theta\sin\theta}{a\theta\cos\theta}
\displaystyle =\frac{\sin\theta}{\cos\theta}=\tan\theta
Hence,
\boxed{\displaystyle \frac{dy}{dx}=\tan\theta}
Question 11: Functions in Parametric Forms
11. If \displaystyle x=\sqrt{a^{\sin^{-1}t}},\quad y=\sqrt{a^{\cos^{-1}t}}, show that \displaystyle \frac{dy}{dx}=-\frac{y}{x}.
Solution
Given,
\displaystyle x=\sqrt{a^{\sin^{-1}t}}=a^{\frac{1}{2}\sin^{-1}t}
and
\displaystyle y=\sqrt{a^{\cos^{-1}t}}=a^{\frac{1}{2}\cos^{-1}t}
Differentiating y w.r.t. t, we get
\displaystyle \frac{dy}{dt}=a^{\frac{1}{2}\cos^{-1}t}\cdot\ln a\cdot\frac{d}{dt}\left(\frac{\cos^{-1}t}{2}\right)
\displaystyle =a^{\frac{1}{2}\cos^{-1}t}\cdot\ln a\cdot\frac{1}{2}\left(-\frac{1}{\sqrt{1-t^2}}\right)
or
\displaystyle =-\frac{\ln a}{2\sqrt{1-t^2}}\,a^{\frac{1}{2}\cos^{-1}t}
Since \displaystyle y=a^{\frac{1}{2}\cos^{-1}t},
\displaystyle \frac{dy}{dt}=-\frac{y\ln a}{2\sqrt{1-t^2}} \qquad ...(1)
Differentiating x w.r.t. t, we get
\displaystyle \frac{dx}{dt}=a^{\frac{1}{2}\sin^{-1}t}\cdot\ln a\cdot\frac{d}{dt}\left(\frac{\sin^{-1}t}{2}\right)
\displaystyle =a^{\frac{1}{2}\sin^{-1}t}\cdot\ln a\cdot\frac{1}{2}\left(\frac{1}{\sqrt{1-t^2}}\right)
or
\displaystyle =\frac{\ln a}{2\sqrt{1-t^2}}\,a^{\frac{1}{2}\sin^{-1}t}
Since \displaystyle x=a^{\frac{1}{2}\sin^{-1}t},
\displaystyle \frac{dx}{dt}=\frac{x\ln a}{2\sqrt{1-t^2}} \qquad ...(2)
Using the formula
\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}
Dividing (1) by (2), we get
\displaystyle \frac{dy}{dx}=\frac{-\frac{y\ln a}{2\sqrt{1-t^2}}}{\frac{x\ln a}{2\sqrt{1-t^2}}}
\displaystyle =-\frac{y}{x}
Hence,
\boxed{\displaystyle \frac{dy}{dx}=-\frac{y}{x}}
Common Mistakes to Avoid
- Reversing the formula: A very common mistake is to write \displaystyle \frac{dy}{dx}=\frac{\frac{dx}{dt}}{\frac{dy}{dt}}. Always remember that \displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}.
- Not differentiating both variables with respect to the parameter: In parametric differentiation, first find \frac{dy}{dt} and \frac{dx}{dt}. Do not attempt to differentiate y directly with respect to x.
- Ignoring the parameter: The parameter may be t, \theta or any other variable. The method remains the same irrespective of the symbol used.
- Making errors while differentiating trigonometric functions: Remember that \frac{d}{dt}(\sin t)=\cos t, \frac{d}{dt}(\cos t)=-\sin t, \frac{d}{dt}(\tan t)=\sec^2 t and \frac{d}{dt}(\sec t)=\sec t\tan t.
- Forgetting the Product Rule: Expressions such as \theta\sin\theta and \theta\cos\theta require the Product Rule. Differentiate both factors separately.
- Making sign errors: Negative signs are frequently missed while differentiating functions such as \cos\theta, \cos2t and \cos^{-1}t.
- Using multiple-angle identities incorrectly: Identities such as \sin2t=2\sin t\cos t, \sin3t=3\sin t-4\sin^3 t and \cos3t=4\cos^3 t-3\cos t should be applied carefully during simplification.
- Forgetting special derivatives: For example, \displaystyle \frac{d}{dt}\left(\log\tan\frac{t}{2}\right)=\frac{1}{\sin t}. Such results are often useful in trigonometric parametric equations.
- Not simplifying the final answer: Expressions such as \frac{\sin\theta}{\cos\theta} and \frac{\cos\theta}{\sin\theta} should be simplified to \tan\theta and \cot\theta respectively whenever possible.
- Using an incorrect formula for second derivatives: Remember that
\displaystyle \frac{d^2y}{dx^2}=\frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}
And
\displaystyle \frac{d^2y}{dx^2}\ne\frac{\frac{d^2y}{dt^2}}{\frac{d^2x}{dt^2}}
Continue Learning
After completing Exercise 5.6 of Continuity and Differentiability, you should now be familiar with differentiating functions given in parametric form using \displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}. In Exercise 5.7, we extend this concept further and learn how to find second order derivatives for functions expressed in parametric form.
To strengthen your understanding further, make sure that you revise:
- Meaning of a parameter and parametric equations
- The formula \displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}
- Derivatives of algebraic functions using the Power Rule
- Derivatives of trigonometric functions such as \sin x,\cos x,\tan x,\sec x
- Derivatives of inverse trigonometric functions such as \sin^{-1}x and \cos^{-1}x
- Derivative of exponential functions of the form a^u
- Applying the Chain Rule while differentiating composite functions
- Applying the Product Rule in expressions such as \theta\sin\theta and \theta\cos\theta
- Applying the Quotient Rule in rational and trigonometric expressions
- Standard trigonometric identities including \sin2\theta, \cos2\theta, \sin3\theta and \cos3\theta
- Half-angle identities involving \sin\frac{\theta}{2}, \cos\frac{\theta}{2} and \tan\frac{\theta}{2}
- Simplifying trigonometric expressions before writing the final answer
Explore More
Try solving a variety of parametric differentiation problems on your own. Practise finding \frac{dy}{dt} and \frac{dx}{dt} separately before using the formula \displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}. Also solve questions involving trigonometric, inverse trigonometric, exponential and logarithmic functions given in parametric form. Regular practice will help you apply the Chain Rule, Product Rule and Quotient Rule correctly and simplify the final answer with confidence.
All the best and keep learning 👍
