Higher Order Derivatives of Functions in Parametric Form - Solved Examples

Sometimes relationships between two variables are represented by using a third variable to make things easier to handle. In Mathematics, this third variable is called a parameter. Mostly, curves are represented by the parameters, for example, x = a and y = 2at. For these types of curves, we have to find the derivative as given below:

\(\begin{array}{l}\frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx} = \frac{y’(t)}{x’(t)}\end{array} \)

In this section, we will solve problems of curves which are defined using parameters and explain how their derivatives can be found using parametric differentiation.

Higher-order Derivative in Parametric Form

We know that the derivative of the first-order in parametric equations is given by:

\(\begin{array}{l}\frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{ dx} = \frac{y’(t)}{x’(t)}\end{array} \)

Where t is the parameter.

Then, if we differentiate the given first derivative w.r.t. x, we will get 2nd order derivative as given below:

Solved Examples

Below are a few solved examples on first-order derivatives and higher-order derivatives.

Examples on First Order Derivative

Problem 1: Find dy/dx if y = sin t and x = cos t.

Solution: y = sin t and x = cos t, then

y’(t) = cos t

x’(t) = – sin t

\(\begin{array}{l}\frac{dy}{dx}=\frac{cos(t)}{-sin(t)} \end{array} \)

or

\(\begin{array}{l}\frac{dy}{dx} = – cot \ t\end{array} \)

Problem 2: Find dy/dx if y = sec³A and x = a + tan³A.

Solution: y = sec³A and x = a + tan³A, then

⇒ y’(A) = 3 sec² A.sec A.tan A

And x’(A) = 3 tan² A . sec² A

\(\begin{array}{l}\Rightarrow \frac{dy}{dx} = \frac{3 sec^2 A . sec A . tan A}{3 tan^2 A.sec^2 A}\end{array} \)

\(\begin{array}{l}=\frac{sec A}{tan A} \end{array} \)

\(\begin{array}{l}=\frac{1}{sin A} \end{array} \)

= cosec A

Examples on Higher Order Derivative

Problem 3: Find the 2nd order derivative if x = p(A – sin A) and y = p(1 – cos A).

Solution: x = p(A – sin A) and y = p(1 – cos A), then

⇒ y’(A) = dy/dA = p sin A

⇒ x’(A) = dx/dA = p(1-cos A)

\(\begin{array}{l}\Rightarrow \frac{dy}{dx} = \frac{p sin A}{p(1-cos A)}\end{array} \)

\(\begin{array}{l}=\frac{2 sin \frac{A}{2} cos \frac{A}{2}}{2sin^2 \frac{A}{2}}\end{array} \)

\(\begin{array}{l}=\frac{cos \frac{A}{2}}{sin^2 \frac{A}{2}}\end{array} \)

\(\begin{array}{l}=cot \frac{A}{2}\end{array} \)

Now,

\(\begin{array}{l}\frac{d}{dA}(\frac{dy}{dx})= \frac{1}{2} \times -cosec^2 (\frac{A}{2})\end{array} \)

and, dx/dA= p(1-cos A)

\(\begin{array}{l}\Rightarrow \frac{dA}{dx} = \frac{1}{p(1-cos A)}\end{array} \)

\(\begin{array}{l}\Rightarrow \frac{d^2 y}{dx^2} = \frac{d}{dA}(\frac{dy}{dx}). \frac{dA}{dx}\end{array} \)

\(\begin{array}{l}\Rightarrow \frac{d^2 y}{dx^2}= \frac{1}{2} \times -cosec^2 (\frac{A}{2}) \times \frac{1}{(p(1-cos A)}\end{array} \)

Problem 4: Find the second order derivative of the Y w..r.t. X, Y = (1 + t²)/t and x = 4t + t³.

Solution: Y = (1 + t²)/t and x = 4t + t³, then

Parametric Derivative Equations Example

Methods of Differentiation – JEE Solved Questions

Important Theorems of Differentiation for JEE

Frequently Asked Questions

Q1

What do you mean by the derivative of a function?

The derivative of a function is the rate of change of a function with respect to a point lying in its domain.

Q2

How do you denote the first-order and second-order derivative of a function f(x)?

The first-order derivative of f(x) is denoted by f’(x), and the second-order derivative of f(x) is denoted by f’’(x).

Q3

Give the equation for the derivative of the first order in parametric form.

The derivative of the first order in parametric form is given by dy/dx = (dy/dt) × (dt/dx), where t is the parameter.

Q4

Give the equation for the derivative of the second order in parametric form.

The derivative of the second order in parametric form is given by d²y/dx² = (d/dx) (dy/dx) = (d/dt)((dy/dt) × (dt/dx))× (dt/dx), where t is the parameter.

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