# Prov (fgh)' = f'gh + fg'h +fgh'

zketrouble

## Homework Statement

Hi folks,

I've been trying to teach myself Calculus and I found a problem on the net which asks to prove that (fgh)' = f'gh +fg'h +fgh'. I think I'm making this problem a lot more complicated than it needs to be.

## The Attempt at a Solution

Here's my attempt, the first part was all brainstorming:

d/dx 7x^2 = 14x
d/dx 7x^2; f(x) = 7 and g(x)=x^2

...f'g = (0)x^2; fg' = 7(2x)
0x^2 + 14x = 14x

Now lets try (constant)*x*x^n

k*x*x^n = kx^(n+1)

If: d/dx (fgh)' = f'gh + fg'h + fgh'
Then: d/dx k*x*x^n = d/dx kx^(n+1)

d/dx (k*x*x^n) = k'(x)(x^n) + k(x)'(x^n) + k(x)(x^n)'
d/dx (k*x*x^n) = 0(x)(x^n) + k(1)(x^n) + k(x)[nx^(n-1)]
d/dx (k*x*x^n) = 0 + k(x^n) + kx[nx^(n-1)]
d/dx (k*x*x^n) = kx^n + knx^n
d/dx (K*x*x^n) = 2kn(x^n)

d/dx [kx^(n+1)] = (n+1)kx^n
d/dx [kx^(n+1)] = nkx^n + kx^n
d/dx [kx^(n+1)] = 2kn(x^n)

I'm not sure if this thoroughly proves the statement (fgh)' = f'gh + fg'h + fgh' because it fails to take into account varying coefficients on terms of varying degree. So I tried again using a similar strategy, this time failing:

Let (a,b,c) = three coefficient constants. Let (d,e,f) = three exponential constants. When I use the letter e I am NOT referring to the number 2.71828.

(ax^d)*(bx^e)*(cx^f) = abcx^(d+e+f)
d/dx (ax^d)*(bx^e)*(cx^f) = d/dx abcx^(d+e+f)

Letting the three terms on the left-hand of the equation be denoted as functions F, G, and H,

If: (FGH)' = F'GH + FG'H + FGH'
Then: d/dx (ax^d)*(bx^e)*(cx^f) = d/dx abcx^(d+e+f)
and the triple product rule should be consistent with the equation, and the two derivatives below should obtain the same result.

d/dx (ax^d)*(bx^e)*(cx^f)

= adx^(d-1)*(bx^e)(cx^f) + bex^(e-1)*(ax^d)(cx^f) + cfx^(f-1)*(ax^d)(bx^e)
= adx^(d-1)*bcx^(e+f+1) + bex^(e-1)*acx^(d+f+1) + cfx^(f-1)*abx^(d+e+1)
= abcdx^(d+e+f) + abcex^(d+e+f) + abcfx^(d+e+f)
= abc[dx^(d+e+f) + ex^(d+e+f) + fx^(d+e+f)]

d/dx abcx^(d+e+f)

= abc(d+e+f)x^(d+e+f-1)
= abcdx^(d+e+f-1) + abcex^(d+e+f-1) + abcfx^(d+e+f-1)
= abc[dx^(d+e+f-1) + ex^(d+e+f-1) + fx^(d+e+f-1)

abc[dx^(d+e+f-1) + ex^(d+e+f-1) + fx^(d+e+f-1)

does NOT = abc[dx^(d+e+f) + ex^(d+e+f) + fx^(d+e+f)]

So I've either disproven one of the fundamental laws of differentiation (unlikely) or made an error somewhere (much more likely). Can anybody see where the error is, and does anybody have a better idea on how to solve the problem?

Thanks!!

Homework Helper
Gold Member

## Homework Statement

Hi folks,

I've been trying to teach myself Calculus and I found a problem on the net which asks to prove that (fgh)' = f'gh +fg'h +fgh'. I think I'm making this problem a lot more complicated than it needs to be.

...

So I've either disproven one of the fundamental laws of differentiation (unlikely) or made an error somewhere (much more likely). Can anybody see where the error is, and does anybody have a better idea on how to solve the problem?

Thanks!!

Apply (fg)' = f'g + g'f to your problem written as

$$(f\cdot g\cdot h)' = (f\cdot(g\cdot h))'$$

and expand it.