[tex]f(x)=\ln x\\\\g(x)=\ln11x=\ln11+\ln x\ \ (\ln11=const.)\\\\h(x)=\ln x^2=2\ln x\\\\p(x)=\ln3x^2=\ln 3+\ln x^2=\ln3+2\ln x\ \ (\ln3=const.)[/tex]
Used:
[tex]\ln a+\ln b=\ln(ab)\\\\\ln a^n=n\log a[/tex]
Functions have the same derivatives if they differ a constant.
h(x) = f(x); g(x) = f(x) + const.
h'(x) = f'(x)
g'(x) = (f(x) + const.)' = f'(x) + (const.)' = f'(x) + 0 = f'(x)
h'(x) = g'(x)
Therefore yuor answer is A. f'(x)=g'(x)
[tex]f'(x)=(\ln x)'=\dfrac{1}{x}\\\\g'(x)=(\ln11x)=\dfrac{1}{11x}\cdot11=\dfrac{1}{x}[/tex]
Used:
[tex](\ln x)'=\dfrac{1}{x}\\\\f'(g(x))=f'(g(x))\cdot g'(x)[/tex]
