# Higher-Order Derivative of Functions In Parametric Form

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## 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.