Derivative of Functions in Parametric Forms

Parametric Differentiation refers to the differentiation of a function in which the dependent and independent variables are equated to a third variable. Derivatives of the functions express the rate of change in the functions. We know how to calculate the derivatives for standard functions. Chain rule, Product rule, and Quotient rule are used to calculate the derivatives of the complex functions which are made up of composition from two or more functions. These functions have two variables that are related to each other in an implicit or explicit manner.

Sometimes we encounter functions in which variables are not related to each other implicitly or explicitly, instead, they are related to each other through a third variable. In this case, we use Parametric Differentiation, which we study in Class 12 for the first time. This article helps in the introduction of the topic to everyone. Here we discussed Parametric Differentiation in detail including methods to find the derivative of a function in parametric form, as well as various other solved examples.

What is Parametric Differentiation?

Parametric Differentiation or Derivative of Function in Parametric Form is the process of finding the derivative of a function in which the dependent variable 'y' and the independent variable 'x' are equated to a third independent variable 't'.

Most often we come across a function that is given straightforwardly manner that y is directly dependent on x i.e., y = f(x) so finding the derivative becomes easy and we easily differentiate y with respect to x and calculate dy/dx. Sometimes we have a function in implicit form i.e. both x and y variables are present in the function and the function is given as f(x, y) in this case we either convert the function into explicit form or apply implicit differentiation.

In Parametric Differentiation, since the variables are connected to each other via a third variable we first differentiate the variables x and y with respect to the third independent variable and their find the ratio of their derivative.

Parametric Equation

Parametric Equation is a equation in which the two variables are equated to a third variable usually denoted by 't'. This third variable is called parameter and hence the name is Parametric Equation. This type of equation is usually used in replacing the coordinates from the cartesian form of equation of a curve and convert it into a variable form called as parametric form.

In Parametric Function the two variables are given as x = f(t) and y = g(t) i.e. both are a function of another independent variable 't'.

How to Find Derivatives of Functions in Parametric Forms?

Let's say we have two variables x and y, usually, such variables are related to each other in an implicit or an explicit manner. But in some cases, these variables are related to each other through a third variable. This form is called the parametric form of the equation and the variable is called a parameter. We have x = f(t) and y = g(t) here, t is a parameter.

We will first differentiate x and y with respect to 't' separatly. On differentiating x with respect to 't' we get dx/dt and on differentiating y by 't' we get dy/dt.

Now to the derivative of y with respect to x is given by the formula dy/dx = (dy/dt).(dt/dx). The formula for parametric differentiation is also expressed in the image attached below:

The derivative of such functions is given by chain rule,

Using Chain Rule, dy/dt can be written as

\frac{dy}{dt} = \frac{dy}{dx}\cdot \frac{dx}{dt}

Now, rearranging the terms, we get

\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} , where\frac{dx}{dt} \ne 0

Thus,\frac{dy}{dx}= \frac{g'(t)}{f'(t)}  [as\frac{dy}{dx} = g'(t) and\frac{dx}{dt} = f'(t)  ]

Proof of Parametric Differentiation Formula

Since y and x are dependent on 't', then any change in 't' would also cause a change in 'y' and 'x'. Hence, for small change in 't' given as Δt the corresponding changes in x and y are Δx and Δy.

We can write Δy/Δx = (Δy/Δt)/(Δx/Δt)

Taking limit on both sides

lim _Δx→0 Δy/Δx = lim _Δt→0 (Δy/Δt)/lim _Δt→0 (Δx/Δt)

Using the Concept of Limit and Derivatives, we have

dy/dx = (dy/dt)/(dx/dt)

Read More,

• Derivative of Inverse Trigonometric Function
• Chain Rule
• Application of Derivatives

Parametric Differentiation Examples

Example 1: Find dy/dx, if x = acos(θ) , y = asin(θ).

Solution:

x = acos(θ) and y = asin(θ)

dx/dθ ​= −sin(θ)

dy/dθ = a cosθ

Now, let's find out

dy/dx = (dy/dt)/(dx/dt)

⇒ dy/dx = (dy/dθ)/(dx/dθ)

⇒ dy/dx = acosθ/-asinθ

⇒ dy/dx = -cot θ

Example 2: Find dy/dx, if x = acos²(θ) , y = asin²(θ).

Solution:

x = acos²(θ) , y = asin²(θ)

dx/dθ = -2acosθsinθ

dy/dθ = 2acosθsinθ

Now, let's find out

dy/dx = (dy/dt)/(dx/dt)

⇒ dy/dx = (dy/dθ)/(dx/dθ)

⇒\frac{dy}{dx} = \frac{-2acos(\theta)sin(\theta)}{2acos(\theta)sin(\theta)}

⇒ dy/dx = -1

Example 3: Find dy/dx, if x = at² + 2t, y = t at t = 0.

Solution:

x = at² + 2t, y = t

dx/dt = 2at + 2

dy/dt = 2

Now, let's find out

dy/dx = (dy/dt)/(dx/dt)

⇒\frac{dy}{dx} = \frac{2}{2at + 2}

⇒ dy/dx = 2/2

⇒ dy/dx = 1

Example 4: Find dy/dx, if x = at³ + 2t², y = t² at t = 1.

Solution:

x = at³ + 2t², y = t²

dx/dt = 3at² + 4t

dy/dt = 2t

Now, let's find out

dy/dx = (dy/dt)/(dx/dt)

⇒\frac{dy}{dx} = \frac{2t}{3at^2 + 4}

⇒\frac{dy}{dx} = \frac{2}{3a + 4}

Example 5: Find dy/dx, if x = 4t, y = 1/t at t = 1.

Solution:

x = 4t, y = 1/t

dx/dt = 4

dy/dt = -1/t²

Now, let's find out

dy/dx = (dy/dt)/(dx/dt)

⇒\frac{dy}{dx} = \frac{-\frac{1}{t^2}}{4}

⇒ dy/dx = -1/4t²

At t = 1

dy/dx = -1/4

Example 6: Find dy/dx, if x = 4e^t, y = cos(t)at t = 1.

Solution:

x = 4e^t, y =cos(t)

dx/dt = 4et

dy/dt = -sin(t)

Now, let's find out\

dy/dx = (dy/dt)/(dx/dt)

⇒\frac{dy}{dx} = \frac{-sin(t)}{4e^t}

At t = 1

⇒\frac{dy}{dx} = -\frac{sin(1)}{4e}

Example 7: Find dy/dx, if x = e^t + sin(t), y = t², at = 0.

Solution:

x = e^t + sin(t), y =t²

dx/dt = e^t + cos(t)

dy/dt = 2t

Now, let's find out

dy/dx = (dy/dt)/(dx/dt)

⇒\frac{dy}{dx} = \frac{2t}{e^t + cos(t)}

At t = 0

⇒ dy/dx = 0

Example 8: Find dy/dx , if x = tsin(t), y = cos(t), at t = π/2

Solution:

x = tsin(t), y = cos(t)

dx/dt = t.cos(t) + sin(t)

dy/dt = -sin(t)

Now, let's find out

dy/dx = (dy/dt)/(dx/dt)

⇒\frac{dy}{dx} = \frac{-sin(t)}{tcos(t) + sin(t)}

At t = π/2

⇒\frac{dy}{dx} = \frac{-sin(\frac{\pi}{2})}{\frac{\pi}{2}cos(\frac{\pi}{2}) + sin(\frac{\pi}{2})}

⇒ dy/dx = -1

Example 9: Find dy/dx if x = 4t² + 10, y =t²at t = 1.

Solution:

x = 4t² + 10, y =t²

dx/dt = 8t

dy/dt = 2t

Now, let's find out

dy/dx = (dy/dt)/(dx/dt)

⇒ dy/dx = 2t/8t

At t = 1

dydx = 1/4

Example 10: Find dy/dx if x = at², y =2at at t = 1.

Solution:

x = at², y =2at

dx/dt = 2at

dy/dt = 2a

Now, let's find out

dy/dx = (dy/dt)/(dx/dt)

dy/dx = 2a/2at

At t = 1

dy/dx = 1

Practice Questions on Parametric Differentiation

Q1: Find dy/dx if x = sin²t and y = cos³t

Q2: Find dy/dx if x = at³ and y = 3at

Q3: Find dy/dx if x = 3 log t and y = e^2t

Q4: Find dy/dx if x = a sec θ and y = b tan θ