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## 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.

## Homework Equations

## 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 let's 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!