examples of PRODUCT and QUOTIENT derivatives?

Question

pls write the reference books/sites used

Answer 1

What do you mean by this? Derivative of a product and quotient of two or more functions?

if h(x)=f(x)*g(x)
=> h'(x)=f(x)*g'(x)+g(x)*f(x)

for example:
if h(x)=(x^3+3x^2+4)*(x^5-x^4)
if we let f(x)=x^3+3x^2+4 and g(x)=x^5-x^4
=> f'(x)=3x^2+6x and g'(x)=5x^4-4x^3
hence, h'(x)=(x^3+3x^2+4)*(5x^4-4x^3)+(x^5-x^4)*(3x^2+6x)

for quotient rule
if h(x)=f(x)/g(x)
=> h'(x)=[g(x)*f'(x)-f(x)*g'(x)]/[g(x)]^2

for example
if h(x)=(x^3+1)/(x^2-4)
if we let f(x)=x^3+1 and g(x)=x^2-4
=> f'(x)=3x^2 and g'(x)=2x
thus, h'(x)=[(x^2-4)*(3x^2)-(x^3+1)*(2x)]/[x^2-4]^2


if you a the proof for these formulas, there are calculus books available.

Answer 2

Chain rule must be invoked for a derivative of products.

Example 1:

Find derivative of: ( x² + 3x ) ( 5x + 3 )

First, notice that this is a product, and so we need to use the product rule:

d/dx ( u * v ) = uv ' + vu'

So in Example 1: Let u = (x² + 3x)
............................ Let v = (5x + 3)

calculate u ' and v ' now...

u' = d/dx u = d/dx (x² + 3x ) = 2x + 3
v' = d/dx v = d/dx (5x + 3 ) = 5

Put together the components so that: uv' + vu'

uv' + vu' = (x² + 3x)(5) + (5x + 3)(2x + 3)

Now we have to simplify...

( 5x² + 15x ) + (10x² +15x +6x + 9)


Collect Like Terms ...

15x² + 36x + 9 is your final simlified answer!

Answer 3

d/dx (f(x)*g(x))
[h→0]lim (f(x+h)g(x+h) - f(x)g(x))/h
[h→0]lim ((f(x) + f(x+h) - f(x))g(x+h) - f(x)g(x))/h
[h→0]lim (f(x)(g(x+h) - g(x)) + (f(x+h) - f(x))g(x+h))/h
[h→0]lim (f(x)(g(x+h) - g(x))/h + ((f(x+h) - f(x))g(x+h))/h
f(x) [h→0]lim (g(x+h)-g(x))/h + [h→0]lim ((f(x+h) - f(x))g(x+h))/h
f(x)g'(x) + [h→0]lim g(x+h)(f(x+h) - f(x))/h
f(x)g'(x) + [h→0]lim g(x+h) * [h→0]lim (f(x+h) - f(x))/h
f(x)g'(x) + g(x)f'(x)

Alternatively, the intutive version:

[Δx→0]lim Δ(fg)/Δx
[Δx→0]lim ((f+Δf)(g+Δg) - fg)/Δx
[Δx→0]lim (fg + Δfg + fΔg + ΔfΔg - fg)/Δx
[Δx→0]lim gΔf/Δx + fΔg/Δx + ΔfΔg/Δx
gf' + fg' + g'[Δx→0]lim Δf
gf'+fg'

Some rigor is lost in the intuitive version, but it makes it much clearer where the product rule comes from.

The quotient rule can be derived from the product rule thus:

Let h(x)=f(x)/g(x)
g(x)h(x)=f(x)
Taking the derivative of both sides:
g'(x)h(x) + g(x)h'(x) = f'(x)
g(x)h'(x) = f'(x) - g'(x)h(x)
g(x)h'(x) = f'(x) - g'(x)f(x)/g(x)
g(x)h'(x) = f'(x)g(x)/g(x) - f(x)g'(x)/g(x)
g(x)h'(x) = (f'(x)g(x) - f(x)g'(x))/g(x)
h'(x) = (f'(x)g(x) - f(x)g'(x))/g(x)²

It may also be proved directly, if desired (proof left as an exercise to the reader).

Answer 4

I dont know