Class 12 Maths Continuity Differentiability Derivatives of Functions in Parametric Forms

Derivatives of Functions in Parametric Forms

When a relationship between 2 variables is established with a link of third variable, then this third variable is called parameter.

Suppose x = f(t), y = g(t)

This relationship between x and y is said to be of parametric form with t as a parameter.

To find the derivative of a function in this form, we use the chain rule.

i.e dy/dx = dy/dt * dt/dx

=> dy/dx = (dy/dt)/(dx/dt), where dx/dt ≠ 0

=> dy/dx = [d{g(t)}/dt]/ [d{f(t)}/dt]

=> dy/dx = g’(t)/f’(t)

Here d{g(t)}/dt] = g’(t), [d{f(t)}/dt = f’(t) and f’(t) ≠ 0

Problem: Find the derivative of the function x = sin t, y = cos 2t

Solution:

The given equations are: x = sin t, y = b cos 2t

Now, dx/dt = d(sin t)/dt

=> dx/dt = cos t

and dy/dt = d(cos 2t)/dt

=> dy/dt = -sin 2t * d(2t)/dt

=> dy/dt = -2 sin 2t

Now, dy/dx = (dy/dt)/(dx/dt)

=> dy/dx = (-2 sin 2t)/cos t

=> dy/dx = (-2 * 2 * sin t * cos t)/cos t

=> dy/dx = -4 sin t

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