Mathematics Derivatives of Functions in Parametric Forms And Second Order Derivative For CBSE-NCERT
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### Topic covered

star Derivatives of Functions in Parametric Forms
star Second Order Derivative :

### Derivatives of Functions in Parametric Forms :

\color{green} ✍️ whenever the relation between two variables is neither explicit nor implicit, We express the relation between them via a third variable. The third variable is called the parameter.

\color{green} ✍️ A relation expressed between two variables x and y in the form x = f(t), y = g (t) is said to be parametric form with t as a parameter.

\color{green} ✍️ In order to find derivative of function in such form, we have by chain rule.

color{red}{(dy)/(dt) =(dy)/(dx) * (dx)/(dt)}

or (dy)/(dx) = ((dy)/(dt))/((dx)/(dt))  (whenever (dx)/(dt) ne 0)

Thus => (dy)/(dx) = (g'(t))/(f'(t))  (as (dy)/(dt) = g'(t) and (dx)/(dt) = f'(t)) [provided f ′(t) ≠ 0]
Q 3115723669

Find  (dy)/(dx) , if x = at^2 , y = 2 at

Class 12 Chapter 5 Example 35
Solution:

Given that x = at^2, y = 2at

So  (dx)/(dt) =2 at  and  (dy)/(dt) = 2a

Therefore  (dy)/(dx) = ( (dy)/(dt) )/( (dx)/(dt) ) = (2a)/(2at) = 1/t
Q 3115823760

Find  (dy)/(dx) , if x =a ( theta + sin theta) , y = a (1- cos theta)

Class 12 Chapter 5 Example 36
Solution:

We have  (dx)/(d theta) = a (1+ cos theta) , (dy)/(d theta) = a (sin theta)

Therefore  (dy)/(dx) = ( (dy)/(d theta) )/( (dx)/(d theta) ) = (a sin theta)/(a (1+ cos theta) ) = tan theta/2
Q 3135234162

Find  (dy)/(dx) , if x^(2/3) + y^(2/3) = a^(2/3)
Class 12 Chapter 5 Example 37
Solution:

Let x = a cos^3 θ, y = a sin^3 θ. Then

x^2/3 + y^2/3 = (a cos^3 θ ) , y = a sin^3 θ. Then

x^(2/3) + y^(2/3) = (a cos^3 θ)^(2/3) + (a sin^3 θ)^(2/3)

= a^(2/3) (cos^2 θ + (sin^2 θ) ) = a^(2/3)

Hence, x = a cos^3 θ, y = a sin^3θ is parametric equation of x^(2/3) + y^(2/3) = a^(2/3)

Now (dx)/(d theta) = -3 a cos^2 theta sin theta  and  (dy)/(d theta) = 3a sin^2 theta cos theta

Therefore  (dy)/(dx) = ( (dy)/(d theta) )/( (dx)/(d theta) ) = (3 a sin^2 theta cos theta )/( -3 a cos^2 theta sin theta) = root (1/3) (y/x)

### Second Order Derivative :

\color{green} ✍️ Let y = f (x). Then (dy)/(dx)= f'(x).....................(1)

\color{green} ✍️ If f ′(x) is differentiable, we may differentiate (1) again w.r.t. x. Then, the left hand side becomes color{red}{d/(dx) ((dy)/(dx))} which is called the second order derivative of y w.r.t. x and is denoted by (d^2y)/(dx^2)

\color{green} ✍️ The second order derivative of f (x) is denoted by f ″(x). It's denoted by D^2 y  or y″ or y^2 if y = f (x). We remark that higher order derivatives may be defined similarly.
Q 3155334264

Find  (d^2 y )/(dx^2) , if  y = x^3 + tan x.
Class 12 Chapter 5 Example 38
Solution:

Given that y = x^3 + tan x. Then

(dy)/(dx) = 3x^2 + sec^2 x

Therefore  (d^2 y)/(dx^2) = d/(dx) (3x^2 + sec^2 x)

= 6x + 2 sec x * sec x tan x = 6x + 2 sec2 x tan x
Q 3115334269

If y = A sin x + B cos x , then prove that  ( d^2 y )/(dx^2 ) + y = 0
Class 12 Chapter 5 Example 39
Solution:

We have

(dy)/(dx) = A cos x – B sin x

and  (d^2 y )/(dx^2) = d/(dx) (A cos x – B sin x)

= – A sin x – B cos x = – y

Hence  (d^2 y )/(dx^2) + y = 0
Q 3125434361

If y = 3e^(2x) + 2e^(3x) , prove that (d^2 y)/(dx^2) -5 (dy)/(dx) + 6 y = 0
Class 12 Chapter 5 Example 40
Solution:

Given that y = 3e^(2x) + 2e^(3x). Then

(dy)/(dx) = 6e^(2x) + 6e^(3x) = 6 (e^(2x) + e^(3x) )

Therefore  (d^2 y )/(dx^2) = 12e^(2x) + 18e^(3x) = 6 (2e^(2x) + 3e^(3x ) )

Hence  (d^2 y )/(dx^2) -5 (dy)/(dx) + 6y = 6 (2e^(2x) + 3 e^(3x) )

– 30 (e^(2x) + e^(3x) ) + 6 (3e^(2x) + 2e^(3x) ) = 0
Q 3135434362

If y = sin^(–1) x, show that (1 – x^2 ) (d^2 y )/(dx^2) - x (dy)/(dx) = 0 .
Class 12 Chapter 5 Example 41
Solution:

We have y = sin^(–1) x. Then

(dy)/(dx) = 1/(sqrt (1-x^2) )

or sqrt ( (1-x^2) ) (dy)/(dx ) =1

So d/(dx) ( sqrt (1-x^2) * (dy)/(dx) ) = 0

or  sqrt ( (1-x^2) ) * (d^2 y )/(dx^2) + (dy)/(dx) * d/(dx) (sqrt (1-x^2) ) = 0

or  sqrt (1-x^2) * (d^2 y)/(dx^2) - (dy)/(dx) * (2x)/( 2 sqrt( 1-x^2) ) = 0

Hence  (1-x^2 ) (d^2 y )/(dx^2) -x (dy)/(dx) = 0`