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In each of the topic, the outline of the thought-process, for that topic, is provided for learners and educators.

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» Differentiation under Function Operations

→ Composite form and Chain rule: given `v(u)` and `u(x)` (i.e. `v(u(x))`) then

`(dv)/(dx) = (dv)/(du)(du)/(dx)`

→ Parametric form : given `g(x)` and `h(x)` then

`(dg)/(dh) = (dg//dx)/(dh//dx)`

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In this page, the derivatives of functions in composite form and parametric forms are discussed.

Starting on "Differentiation of Functions in Different Forms". In this page, the derivatives of functions in composite form and parametric forms are discussed.

Finding derivative of composite function.

Given `v(u)` and `u(x)` find `(dv)/(dx)`

`(dv)/(dx)`

`= lim_(Deltax->0) (v(u(x+Delta x)) -v(u(x)))/(Delta x)`* multiplying and dividing by `u(x+Delta x) - u(x)`*

`=lim_(Deltax->0) [``(v(u(x+Delta x)) -v(u(x)))/(u(x+Delta x) - u(x))`` xx (u(x+Delta x) - u(x))/(Delta x)``]`* Substituting `u(x+Delta x) - u(x) = Delta u`If `Delta x ->0`, then `Delta u ->0` with continuity and differentiability conditions on `v` and `u`. *

`= lim_(Delta u -> 0) (v(u(x+Delta x)) -v(u(x)))/(Delta u)`` lim_(Delta x->0)(u(x+Delta x) - u(x))/(Delta x)`

`=(dv)/(du) (du)/(dx)`

What does the above prove?

- `(dv)/(dx) = (dv)/(du) (du)/(dx)`
- `(dv)/(dx) = (dv)/(du)`
- `(dv)/(dx) = (du)/(dx)`

The answer is "`(dv)/(dx) = (dv)/(du) (du)/(dx)`"

Derivative of composite function can be extended to multi-level functions.

`(df)/(dx) = (df)/(dg)(dg)/(dh)(dh)/(dv)(dv)/(dx)`

This is also called **chain rule of differentiation**.

Intuitive understanding of chain rule :

``

In `v(u(x))`, `v(u)` is called outer-function and `u(x)` is called inner-function. In a composite function,

the change in variable causes a change in inner-function and that change is the rate of change of inner-function.

the change in inner-function, in turn, causes change in outer-function. This change, rate of change of outer-function, is with respect to the inner-function.

Thus rate of change of outer function with respect to variable `=` rate of change of inner-function to the variable `xx` rate of change of outer-function to the inner-function.

Finding derivative of function given in parametric form.

`y=f(r)` and `x=g(r)`

`(dy)/(dx) = ?`

`(dy)/(dx)` `= lim_(Delta x -> 0) [ Delta y -: Delta x ]`* Note that change in `x` will reflect as change in `r`. `Delta x = g(r+Delta r) -g(r)` `Delta y = f(r+Delta r) - f(r)`*

`(df)/(dg)`

`= lim_(Delta r -> 0) [``(f(r+Delta r) - f(r))`` -:(g(r+Delta r) -g(r))]`

`= lim_(Delta r -> 0) [``(f(r+Delta r) - f(r))//Delta r`` -: (g(r+Delta r) -g(r))//Delta r``]`

`= lim_(Delta r -> 0) [``(f(r+Delta r) - f(r))//Delta r]`` -: lim_(Delta r -> 0)[(g(r+Delta r) -g(r))//Delta r``]`

`= (df//dr) / (dg//dr)`

`= (dy//dr) / (dx//dr)`

What does the above prove?

- `(dv)/(du) = dv//dx -: du//dx`
- `(dv)/(du) = dv//dx xx du//dx`

The answer is "`(dv)/(du) = dv//dx -: du//dx`"

Intuitive understanding of `(dv)/(du) = (dv//dx) / (du//dx)` Rate of change of a function `v` with respect to another function `u` can be computed by the change in the common variable `x`. The change in the common variable `x` causes change in `v` as well as in `u`. This leads to the given formula.

*comprehensive information for quick review*

Jogger

*comprehensive information for quick review*

Jogger

dummy

**Derivative of Composite Function** : `(dv)/(dx) = (dv)/(du) (du)/(dx)`

**Derivative of function in parametric form: ** `(dv)/(du) = (dv//dx) / (du//dx)`

*practice questions to master the knowledge*

Exercise

*practice questions to master the knowledge*

Exercise

Given `y= sin (x^2)`, what is `(dy)/(dx)`?

Note: `d/(dx) x^2 = 2x`

`d/(dx) sin x = cos x`

- `2x cos(x^2)`
- `x^2cosx+2xsinx`

The answer is "`2x cos(x^2)`"

`(dy)/(dx)`* taking `u=x^2`*

`=(d)/(du) sin u xx d/(dx) x^2`* applying law of derivative*

`=cos u xx 2x`

`=2x cos(x^2)`

Given `y= sin r` and `x=r^2`, what is `(dy)/(dx)`?

Note: `d/(dp) p^2 = 2p`

`d/(dp) sin p = cos p`?

- `2r cos(r)`
- `(cos r) / (2r)`

The answer is "`(cos r) / (2r)`"

`(dy)/(dx)`

`(dy//dr)/(dx//dr)`

`=(cos r)/(2r)`

as `(dy)/(dr) = cos r` and `(dx)/(dr)=2r`

*your progress details*

Progress

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Progress

Finding derivative of composite function is given by first principles. What does the above prove.

1

2

3

The answer is "d v by d x = d v by d u multiplied d u by d x"

Derivative of composite function can be extended to multi-level functions. ;; This is also called chain rule of differentiation.

Intuitive understanding of chain rule. In v of u of x, v of u is called outer function, and u of x is called inner function. In a composite function, the change in variable causes a change in inner-function and that change is the rate of change of inner-function.;; the change in inner-function, in turn, causes change in outer-function. This change, rate of change of outer-function, is with respect to the inner-function.;; Thus rate of change of outer function with respect to variable, =, rate of change of inner-function to the variable, multiplied by, rate of change of outer-function to the inner-function.

Given y = sine of x squared what is d y by d x.

1

2

The answer is " 2 x cos x squared"

Derivative of Composite Function: d v by d x = d v by d u multiplied d u by d x

Finding derivative of function in parametric form is given in first principles. What does the above prove.

1

2

The answer is "d v by d u = d v by d x, divided by , d u by d x"

Given y = sine of r and x = r squared what is d y by d x.

1

2

The answer is " cos r by 2 r"

intuitive understanding of d v by d u = d v by d x, divided by , d u by d x. Rate of change of a function v with respect to another function u can be computed by, the change in the common variable x.;; The change in the common variable x causes change in v as well as in u. This leads to the given formula.

Derivative of function in parametric form is d v by d u = d v by d x, divided by , d u by d x