Simple but nice

We use logarithmic differentiation: take log to both sides of the equation, then differentiate them with respect to $x$ .

$\begin{aligned} \ln F(x) &=& \ln(f(x)) + \ln(g(x)) + \ln(h(x)) \\ \frac { F'(x) }{F(x) } &=& \frac { f'(x) }{f(x) } + \frac { g'(x) }{g(x) } + \frac { h'(x) }{h(x) } \\ \frac { F'(a) }{F(a) } &=& \frac { f'(a) }{f(a) } + \frac { g'(a) }{g(a) } + \frac { h'(a) }{h(a) } \\ 21 &=& 4 - 7 + k \\ k &=& \boxed{24} \end{aligned}$

$(uvw)'=u'vw+uv'w+uvw'$

$F'(a)=f'(a)g(a)h(a)+f(a)g'(a)h(a)+f(a)g(a)h'(a)$

Substituting $f'(a), g'(a), h'(a)$ ,

$21F(a)=f(a)g(a)h(a)[4-7+h]$ $21=h-3\Rightarrow h=24$

F(x) = f(x)g(x)h(x)

F'(x) = f'(x)g(x)h(x) + f(x).( g'(x)h(x) + g(x)h'(x) )

21F(x) = 4f(x)g(x)h(x) + f(x) ( -7g(x)h(x) + k.g(x)h(x) )

21F(x) = ( 4 - 7 + k ).f(x)g(x)h(x)

21F(x) = ( k - 3 )F(x)

k - 3 = 21

k = 24