Higher-Order Derivative of Functions In Parametric Form

Parametric Form

The differentiation of a dependent variable with respect to another dependent variable when a third independent variable is given is termed as parametric differentiation. The third variable is termed as a parameter. Parametric differentiation is used for curves like an ellipse, parabola, hyperbola etc.

If x = ф(t) and y = Ψ(t), then the differentiation of y with respect to x is given by,

\[\frac{dy}{dx}\] = \[\frac{dy}{dt}\] х \[\frac{dt}{dx}\]

The differentiation of y with respect to t is multiplied by the reciprocal of differentiation of x with respect to t. Here, x and y are dependent variables and t is an independent variable. The above equation is derived using the chain rule for derivatives.

Higher-order Derivative:

If the first derivative dy/dx of a function y = f (x) is also a differentiable function, then it can be further differentiated with respect to x and this derivative is denoted by \[\frac{d^{2}y}{dx^{2}}\] which is called the second-order derivative of y with respect to x. Further, if \[\frac{d^{2}y}{dx^{2}}\] is also differentiable, then its derivative is called the third derivative of y which is denoted by \[\frac{d^{3}y}{dx^{3}}\]. Similarly, the nth derivative of y is denoted by \[\frac{d^{n}y}{dx^{n}}\]. All these derivatives are called the successive derivatives of y and this process is known as successive differentiation.

The Parametric Form of Higher-order Derivative:

The derivative of the first order in the parametric equation is given by,

\[\frac{dy}{dx}\] = \[\frac{dy}{dt}\] х \[\frac{dt}{dx}\] = \[\frac{y’(t)}{x’(t)}\]

y'(t) is the differentiation of y with respect to t and x'(t) is the differentiation of x with respect to t.

On differentiating this with respect to x, the second-order derivative will be obtained.

\[\frac{d^{2}y}{dx^{2}}\] = \[\frac{d}{dx}\](\[\frac{dy}{dx}\])

Now, substituting \[\frac{dy}{dt}\] х \[\frac{dt}{dx}\] for dy/dx,

\[\frac{d^{2}y}{dx^{2}}\] = \[\frac{d}{dx}\] (\[\frac{dy}{dt}\] х \[\frac{dt}{dx}\] )

Now, writing \[\frac{d}{dx}\] = \[\frac{\frac{d}{dt}}{\frac{dx}{dt}}\],

\[\frac{d^{2}y}{dx^{2}}\] = \[\frac{\frac{d}{dt}(\frac{dy}{dt} \times \frac{dt}{dx})}{\frac{dx}{dt}}\]

Multiplying numerator by reciprocal of the denominator,

\[\frac{d^{2}y}{dx^{2}}\] = \[\frac{d}{dt}\](\[\frac{dy}{dt}\]  х \[\frac{dt}{dx}\]) х \[\frac{dt}{dx}\] 

Therefore,

\[\frac{d^{2}y}{dx^{2}}\] = \[\frac{d}{dt}\](\[\frac{y’(t)}{x’(t)}\]) х \[\frac{dt}{dx}\]

Solved Examples

Example 1:

If x = \[\frac{1-t^{2}}{1+t^{2}}\] and y = \[\frac{2at}{1+t^{2}}\], find dy/dx.

Solution:

Given,

x = \[\frac{1-t^{2}}{1+t^{2}}\] 

Differentiating x with respect to t (using u/v rule),

\[\frac{dx}{dt}\] = \[\frac{(1+t^{2})(0-2t)-(1-t^{2})(0+2t)}{(1+t^{2})^{2}}\]

Simplifying,

\[\frac{dx}{dt}\] = \[\frac{4t}{(1+t^{2})^{2}}\] 

Given,

y = \[\frac{2at}{1+t^{2}}\]

Differentiating y with respect to t (using u/v rule),

\[\frac{dy}{dt}\] = \[\frac{(1+t^{2})2a - 2at(2t)}{(1+t^{2})^{2}}\]

Simplifying,

\[\frac{dy}{dt}\] = \[\frac{2a(1-t^{2})}{(1+t^{2})^{2}}\]

Now, to find dy/dx, the parametric form is used, that is,

\[\frac{dy}{dx}\] = \[\frac{dy}{dt}\] x \[\frac{dt}{dx}\]

Substituting the values,

\[\frac{dy}{dx}\] = \[\frac{dy}{dt}\] x \[\frac{dt}{dx}\]

\[\frac{dy}{dx}\] = \[\frac{2a(1-t^{2})}{(1+t^{2})^{2}}\] x \[\frac{(1+t^{2})^{2})}{4t}\]

Cancelling out common factors and terms,

\[\frac{dy}{dx}\] = \[\frac{\not{2} a(1-t^{2})}{\not{(1+t^{2})^{2}}}\] x \[\frac{\not{(1+t^{2})^{2}}}{_{2}\not{4}t}\]

\[\frac{dy}{dx}\] = \[\frac{a(1-t^{2})}{2t}\]

Example 2:

Find the second derivative of a sin\[^{3}\]t with respect to a cos\[^{3}\]t at t = π/4.

Solution:

Let y = a sin\[^{3}\]t and x = a cos\[^{3}\]t . \[\frac{d^{2}y}{dx^{2}}\] is to be calculated.

First, differentiate y with respect to t.

\[\frac{dy}{dt}\] = 3a sin\[^{2}\]t cos t

Now, differentiate x with respect to t.

\[\frac{dx}{dt}\] = 3a cos\[^{2}\]t(-sin t)

To find dy/dx, the parametric form is used, that is,

\[\frac{dy}{dx}\] = \[\frac{dy}{dt}\] x \[\frac{dt}{dx}\]

Substituting the values,

\[\frac{dy}{dx}\] = \[\frac{3a sin^{2}t cos t}{3a cos^{2}t (-sin t)}\]

Cancelling out common terms and factors,

\[\frac{dy}{dx}\] = \[\frac{\not{3}a sin^{2}t cos t}{\not{3}a cos^{2}t(-sin t)}\]

\[\frac{dy}{dx}\] = - \[\frac{sin t}{cos t}\]

Since, \[\frac{sin x}{cos x}\] = tan x, therefore,

\[\frac{dy}{dx}\] = - tan t

To find the second-order derivative, dy/dx is differentiated again with respect to x,

\[\frac{d^{2}y}{dx^{2}}\] = - sec\[^{2}\] t \[\frac{dt}{dx}\]

Now, since dt/dx is the reciprocal of dx/dt, substituting the reciprocal of dx/dt in the above expression,

\[\frac{d^{2}y}{dx^{2}}\] = \[\frac{-sec^{2}t}{3a cos^{2}t(-sin t)}\]

Since, sec x = \[\frac{1}{cos x}\] ,

\[\frac{d^{2}y}{dx^{2}}\] = \[\frac{1}{3a}\](\[\frac{sec^{4}t}{sin t}\])

Now substituting π/4 for t,

\[\frac{d^{2}y}{dx^{2}}\] = \[\frac{1}{3a}\](\[\frac{sec^{4} \pi / 4 }{sin \pi / 4}\])

\[\frac{d^{2}y}{dx^{2}}\] = \[\frac{1}{3a}\](\[\frac{4}{1 / \sqrt{2}}\]) 

\[\frac{d^{2}y}{dx^{2}}\] = \[\frac{4\sqrt{2}}{3a}\]

Example 3:

If x = a(t-sin t) and y = a(1-cos t). Find dy/dx. Also, find the second-order derivative.

Solution:

Given,

x = a(t - sin t)

Differentiating x with respect to t,

\[\frac{dx}{dt}\] = a(1 - cos t)

Given,

y = a(1 - cos t)

Differentiating y with respect to t,

\[\frac{dy}{dt}\] = a sin t

To find dy/dx, parametric form is used, that is,

\[\frac{dy}{dx}\] = \[\frac{dy}{dt}\] х \[\frac{dt}{dx}\] 

Substituting the values,

\[\frac{dy}{dx}\] = \[\frac{a sin t}{a(1-cost)}\]

Cancelling out the common term from numerator and denominator,

\[\frac{dy}{dx}\] = \[\frac{sin t}{1-cos t}\] 

Substituting, sin t = 2sin\[\frac{t}{2}\]cos\[\frac{t}{2}\] and 1-cos t = 2 sin\[^{2}\] \[\frac{t}{2}\],

\[\frac{dy}{dx}\] = \[\frac{2sin\frac{t}{2}cos\frac{t}{2}}{2sin^{2}\frac{t}{2}}\]

Simplifying,

\[\frac{dy}{dx}\] = \[\frac{cos \frac{t}{2}}{sin \frac{t}{2}}\]

Since, \[\frac{cos x}{sin x}\] = cot x , therefore,

\[\frac{dy}{dx}\] = cot \[\frac{t}{2}\]

To find second-order derivative, dy/dx is differentiated again with respect to x,

\[\frac{d^{2}y}{dx^{2}}\] = \[\frac{1}{2}\] x -cosec\[^{2}\] \[\frac{t}{2}\] x \[\frac{dt}{dx}\]

Now, since dt/dx is the reciprocal of dx/dt, substituting the reciprocal of dx/dt in the above expression,

\[\frac{d^{2}y}{dx^{2}}\] = \[\frac{1}{2}\] x -cosec\[^{2}\] \[\frac{t}{2}\] x \[\frac{1}{a(1-cos t)}\] 

Example 4:

Find the second derivative of x = ln t and y = t\[^{3}\] + 1.

Solution:

Given,

x = ln t

Differentiating x with respect to t,

\[\frac{dx}{dt}\] = \[\frac{1}{t}\]

Given,

y = t\[^{3}\] + 1

Differentiating y with respect to t,

\[\frac{dy}{dt}\] = 3t\[^{2}\]

To find dy/dx, parametric form is used, that is,

\[\frac{dy}{dx}\] = \[\frac{dy}{dt}\] x \[\frac{dt}{dx}\] 

Substituting the values,

\[\frac{dy}{dx}\] = 3t\[^{2}\] x t

\[\frac{dy}{dx}\] = 3t\[^{3}\] 

To find second-order derivative, dy/dx is differentiated again with respect to x,

\[\frac{d^{2}y}{dx^{2}}\]  = 9t\[^{2}\] x \[\frac{dt}{dx}\]

Now, since dt/dx is the reciprocal of dx/dt, substituting the reciprocal of dx/dt in the above expression,

\[\frac{d^{2}y}{dx^{2}}\]  = 9t\[^{2}\] x t

\[\frac{d^{2}y}{dx^{2}}\]  = 9t\[^{3}\]

Since, x = ln t, therefore,

t = e\[^{x}\]

Substituting this value in \[\frac{d^{2}y}{dx^{2}}\],

\[\frac{d^{2}y}{dx^{2}}\] = 9e\[^{3x}\]

Alternative Method:

Differentiating x with respect to t,

x’ = \[\frac{1}{t}\]

Differentiating again,

x’’ = -\[\frac{1}{t^{2}}\]

Differentiating y with respect to t,

y’ = 3t\[^{2}\]

Differentiating again,

y’’ = 6t

Now,

y’’(x) = \[\frac{x’ y’’ - y’ x’’ }{(x’)^{3}}\]

Substituting values in the above equation,

y’’(x) = \[\frac{\frac{1}{t}(6t) - 3t^{2}(-\frac{1}{t^{2}})}{\frac{1}{t^{3}}}\]

y’’(x) = (6 + 3)t\[^{3}\]

y’’(x) = 9t\[^{3}\] 

y’’(x) = 9e\[^{3x}\]

Did you know

• Parametric differentiation is used to derive many formulas of differentiation. For example:

Assuming,

            y = a\[^{x}\]

Taking log on both sides,

y = a\[^{x}\]

log\[_{e}\] y = log\[_{e}\] a\[^{x}\]

Applying the properties of logarithms,

log\[_{e}\] y = log\[_{e}\] a\[^{x}\]

log\[_{e}\] y = x log\[_{e}\] a

Differentiating both sides with respect to x,

\[\frac{d}{dx}\] log\[_{e}\]y = \[\frac{d}{dx}\] x log\[_{e}\]a

\[\frac{1}{y}\] \[\frac{dy}{dx}\] = 1 . log\[_{e}\] a + x . 0

\[\frac{dy}{dx}\] = y log\[_{e}\]a  

\[\frac{dy}{dx}\] = a\[^{x}\] log\[_{e}\] a        [a\[^{x}\] = y]

Therefore, \[\frac{d}{dx}\](a\[^{x}\]) = a\[^{x}\] log\[_{e}\] a 

• If the equation of the curve be given in parametric form say x = f (t) and y = g (t) then,

\[\frac{dy}{dx}\] = \[\frac{dy}{dt}\] / \[\frac{dx}{dt}\] = \[\frac{g’(t)}{f’(t)}\]

             The equation of the tangent at any point t on the curve is given by,

             y - g(t) = \[\frac{g’(t)}{f’(t)}\] [x - f(t)]

• The conversion of an equation to parametric form is called parameterization. For example, consider the equation, y = x\[^{3}\]. Let x = t. Hence, the equation can be described as, y = t\[^{3}\]. This provides greater efficiency when differentiating or integrating a function.

• The parametric equations of a circle x\[^{2}\] + y\[^{2}\] = a\[^{2}\] are x = a cos θ and y = a sin θ. Therefore, the parametric coordinates of any point lying on the circle x\[^{2}\] + y\[^{2}\] = a\[^{2}\] are (a cos θ, a sin θ).

• The parametric equation of a parabola y\[^{2}\] = 4ax are x = at\[^{2}\] and y = 2at where t is the parameter.

• The parametric equations of an ellipse \[\frac{x^{2}}{a^{2}}\] + \[\frac{y^{2}}{b^{2}}\] = 1 are x = a cosθ and y = bsinθ where 0 ≤ θ ≤ 2π is the parameter.