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

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class 12 chapter 5 - continuity & Differentiability

Derivatives of Functions in Parametric Forms And Second Order Derivative For CBSE-NCERT

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

`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`]

`\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.

`\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`

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