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mathsDifferential CalculusAlgebra of Differentiation

Differentiation Under Basic Arithmetic

In this page, the derivatives of functions that are given as arithmetic operation of multiple functions is discussed. The arithmetic operations are addition, subtraction, multiplication, and division.



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Derivative of a scalar multiple of function. Given `v(x)=color(deepskyblue)au(x)`.

`lim_(delta->0) (v(x+delta)-v(x))/delta`

`quad quad = lim_(delta->0) (color(deepskyblue)au(x+delta)-color(deepskyblue)au(x))/delta`

with continuity and differentiability conditions on `u` and `a`
`quad quad = color(deepskyblue)a lim_(delta->0) (u(x+delta)-u(x))/delta`

`quad quad = color(deepskyblue)a d/(dx) u`

what does the above prove?

  • `(a u)′ = a u′`
  • Derivative of a multiple of a function is multiple of the derivative of the function
  • Both the above
  • Both the above

The answer is "both the above"

Intuitive understanding for
`(au)′ = a u′`
rate of change multiplies when the function is multiplied by a constant.

Solved Exercise Problem:

Given `(dy)/(dx) = 2x^2` and `v=y/5`, what is `(dv)/(dx)`?

  • `4x`
  • `2/5 x^2`
  • `2/5 x^2`

The answer is "`2/5 x^2`"

`(dv)/(dx)`
`= d/(dx) y/5`
`= 1/5 d/(dx) y`
`=1/5 xx 2x^2`
`=2/5 x^2`

Derivative of a multiple of a function is multiple of the derivative of the function.

Derivative of a Multiple:
`(au)′ = a u′`

Finding derivative of sum or difference.

`lim_(delta->0) [(u(x+delta)+-v(x+delta))``-(u(x)+-v(x))]//delta`


`quad = lim_(delta->0) [(u(x+delta)-u(x)) ``+- (v(x+delta)-v(x))]//delta`


with continuity and differentiability conditions on `u` and `v`
`quad = lim_(delta->0) (u(x+delta)-u(x))/delta`

`+- lim_(delta->0) (v(x+delta)-v(x))/delta`


`quad quad = d/(dx) u +- d/(dx) v `

What does the above prove? `

  • `(u+v)′ = u′ + v′`
  • `(u+v)′ = u′ + v′`
  • Derivative of a sum or difference is the sum or difference of derivatives.
  • `(u-v)′ = u′ - v′`
  • all the above

The answer is "all the above".

Intuitive understanding for
`(u+-v)′ = u′ +- v′`
rate of change adds (subtracts) when the function is added(subtracted).

Solved Exercise Problem:

Given `(dy)/(dx) = sinx` and `v=y+20`, what is `(dv)/(dx)`?

  • `sin x + 20`
  • `sin x`
  • `sin x`

The answer is "`sin x`"

`(dv)/(dx)`
`=(d(y+20))/dx`
`=(dy)/(dx) + (d(20))/(dx)`
`=sin x + 0`

Derivative of a sum or difference is sum or difference of derivatives.

Derivative of Sum or Difference:
`(u+v)′ = u′ + v′`

`(u-v)′ = u′ - v′`

Finding derivative of product.

`lim_(delta->0) [(u(x+delta) xx v(x+delta))``-(u(x) xx v(x))]//delta`


adding `color(coral)(u(x) xx v(x+delta))``-color(deepskyblue)(u(x) xx v(x+delta))`


`quad quad = lim_(delta->0) [u(x+delta) xx v(x+delta)``- color(deepskyblue)(u(x) xx v(x+delta))``+color(coral)(u(x) xx v(x+delta))``-u(x) xx v(x)]//delta`


with continuity and differentiability conditions on `u` and `v`
`quad quad = lim_(delta->0) [ color(deepskyblue)(v(x+delta)) xx (u(x+delta)``-color(deepskyblue)(u(x)))]//delta`


`quad + lim_(delta->0) [(color(coral)(u(x)) xx (color(coral)(v(x+delta))``-v(x)))]//delta`

`quad quad = v d/(dx) u + u d/(dx) v`

  • `(uv)′=u′v+u v′`
  • `(uv)′=u′v+u v′`
  • `(uv)′=u′v′`

The answer is "`(uv)′=u′v+u v′`"

Intuitive understanding of
`(uv)′=u′v+u v′`

Two functions are multiplied at every point. Rate of change of the product equals sum of

fixing one function `v` as a constant and rate of change of the other function `u′`

plus fixing the other function `u` as a constant and rate of change of the function `v′`

This is compared to `(au)′ = a u′`.

Solved Exercise Problem:

Given `y=x^2 sin x` what is `(dy)/(dx)`?
Note: `d/(dx) x^2 = 2x`
`d/(dx) sin x = cos x`

  • `2xcos x`
  • `2xsinx+x^2cosx`
  • `2xsinx+x^2cosx`

The answer is "`2xsinx+x^2cosx`"

`(dy)/(dx)`
`=(d)/(dx) (x^2 sin x)`
applying product law of derivative `=sin x (d)/(dx)x^2 + x^2 (d)/(dx) sin x`
`=2xsinx+x^2cosx`

Derivative of product of two function is sum of derivatives of the functions scaled by the value of other function

Derivative of Product: `(uv)′=u′v+u v′`

Finding derivative of division.
`lim_(delta->0) [``u(x+delta) -: v(x+delta)``-(u(x) -: v(x))]//delta`

`= lim_(delta->0) {`` 1/(v(x+delta)v(x))``xx [u(x+delta)v(x)`` - u(x)v(x+delta)]//delta }`


adding `color(coral)(u(x) v(x))-color(deepskyblue)(u(x) v(x))`

`= lim_(delta->0) {``1/(v(x+delta)v(x))`` [u(x+delta)v(x) -color(deepskyblue)(u(x) v(x))``- u(x)v(x+delta)+color(coral)(u(x) v(x))]//delta }`


`= lim_(delta->0) {``1/(v(x+delta)v(x))`` [v(x) (u(x+delta) -color(deepskyblue)(u(x)))``- u(x) (v(x+delta)- color(coral)(v(x)))]//delta }`


`= lim_(delta->0) {``1/(v(x+delta)v(x)) xx `` [v(x) (u(x+delta) -color(deepskyblue)(u(x)))]//delta)``- [u(x) (v(x+delta)- color(coral)(v(x)))]//delta }`

with continuity and differentiability conditions on `u` and `v`
`1/(v^2) ``xx {(du)/(dx) v`` - u (dv)/(dx)}`

What does the above prove?

  • `(u/v)′=(u′v-u v′)/(v^2)`
  • `(u/v)′=(u′v-u v′)/(v^2)`
  • `(u/v)′=(u′)/(v′)`

The answer is "`(u/v)′=(u′v-u v′)/(v^2)`"

Intuitive understanding of `(u/v)′=(u′v-u v′)/(v^2)`

This can be rewritten as
`(u/v)′=``(u′ xx (1/v)`` + u xx (-1/v^2) xx v′`

The first part of the result is :
fixing `1/v` as a constant, rate of change of `u` is taken

The second part of the result is :
fixing `u` as a constant, rate of change of `1/v` is taken.
Rate of change of `1/v` is negative of rate of change of `v` and divided by `v^2`.

Solved Exercise Problem:

Given `y= (sin x)/(x^2)` what is `(dy)/(dx)`? Note: `d/(dx) x^2 = 2x`
`d/(dx) sin x = cos x`

  • `-2(cos x)/(x^3)`
  • `(x^2cosx-2xsinx)/(x^4)`
  • `(x^2cosx-2xsinx)/(x^4)`

The answer is "`(x^2cosx-2xsinx)/(x^4)`"

`(dy)/(dx)`
`=(d)/(dx) [(sin x)/(x^2)]`
applying quotient law of derivative `=(x^2 d/(dx) sin x - sin x d/(dx)x^2 )/ (x^4)`
`=(x^2cosx-2xsinx)/(x^4)`

Derivative of Quotient: `(u/v)′=(u′v-u v′)/v^2`

                            
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