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Thought-Process to Discover Knowledge

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mathsDifferential CalculusAlgebra of Differentiation

Differentiation of Functions in Different Forms

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



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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) (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.

Solved Exercise Problem:

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)`
  • `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)`

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

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))]`

dividing numerator and denominator by `Delta r`

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

with continuity and differentiability conditions on `u` and `a`
`= 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 -: du//dx`
  • `(dv)/(du) = dv//dx xx du//dx`

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

Solved Exercise Problem:

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)`
  • `(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`

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.

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

                            
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