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Algebra of Differentiation

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Home

»  Differentiation under Basic Arithmetic Operations

→  (a u)′ = a u′

→  (u+v)′ = u′ + v′

→  (u-v)′ = u′ - v′

→  (uv)′=u′v+u v′

→  (u/v)′=(u′v-u v′)/(v^2)

Differentiation Under Basic Arithmetic

plain and simple summary

nub

plain and simple summary

nub

dummy

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

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

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

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simple steps to build the foundation

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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.

Keep tapping on the content to continue learning.
Starting on "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.

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

The answer is "both the above"

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

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′
• 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′

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′

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′.

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′)

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.

comprehensive information for quick review

Jogger

comprehensive information for quick review

Jogger

dummy

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

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

(u-v)′ = u′ - v′

Derivative of Product: (uv)′=u′v+u v′

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

practice questions to master the knowledge

Exercise

practice questions to master the knowledge

Exercise

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

• 4x
• 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

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

• sin x + 20
• sin x

The answer is "sin x"

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

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

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

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)

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)`

Progress

Progress

Derivative of a scalar multiple of function is given by first principles. What does the above prove.
u;you;whole;prime;equal
a times u, whole prime = a, u prime
derivative;multiple;function
Derivative of a multiple of a function is multiple of the derivative of the function
both;above
Both the above
The answer is "both the above"
Intuitive understanding for the given result is : rate of change multiplies when the function is multiplied by a constant.
Given d y by d x = 2 x squared and v = y by 5, what is d v by d x.
4
4 x
2;5;squared
2 by 5 x squared
The answer is "2 by 5 x squared"
Derivative of a multiple of a function is multiple of the derivative of the function.
Derivative of a Multiple: a u whole prime equals a u prime
Finding derivative of sum or difference is given by first principles. What does the above prove.
plus;equals;+;plus
derivative of; (u plus v) equals derivative of u ;plus; derivative of v
sum;difference
Derivative of a sum or difference is the sum or difference of derivatives.
minus;equals;minus;-
derivative of; (u minus v) equals derivative of u ;minus; derivative of v
all;above
all the above
The answer is "all the above".
Intuitive understanding for u plus or minus v whole prime equals u prime plus or minus v prime. Rate of change adds or subtracts when the function is added or subtracted.
Given d y by d x = sine x and v = y + 20, what is d v by dx.
1
2
Derivative of a sum or difference is sum or difference of derivatives.
Derivative of Sum or Difference: u plus or minus v whole prime equals u prime plus or minus v prime
Finding derivative of product of two functions is given by first principles. What does the above prove.
1
2
The answer is "u v whole prime = u prime v + u v prime"
Intuitive understanding of u v whole prime = u prime v + u v prime. 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 prime; plus; fixing the other function u as a constant and rate of change of the function v prime. ;; This is compared with a u whole prime = a u prime.
Given y = x squared sine x, what is d y by d x.
1
2
The answer is "2 x sine x + x squared cos x"
Derivative of product of two function is sum of derivatives of the functions scaled by the value of other function
Derivative of Product: u v whole prime = u prime v + u v prime.
Finding derivative of division is given by first principles. What does the above prove.
1
2
The answer is " u divided by v whole prime = u prime v minus u v prime divided by v squared"
Intuitive understanding of, u divided by v whole prime = u prime v minus u v prime divided by v squared : This can be rewritten as u divided by v whole prime = u prime multiplied 1 by v, plus, u multiplied, minus 1 by v squared, multiplied v prime. ;; The first part of the result is, fixing 1 by v as 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 by v is taken. Rate of change of 1 by v is negative of rate change of v divided by v squared.
Given y = sine x by x squared, what is d y by d x.
1
2
The answer is "x squared cos x minus 2 x sine x whole divided by x power 4"
Derivative of Quotient: u divided by v whole prime = u prime v minus u v prime divided by v squared

we are not perfect yet...
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