Implicit differentiation to find property of a function Im asked to show that if $h(1)=0$ and $h'(x)={1\over{x}}$ then $a,b&gt;0$ show $h(ab)=h(a)+h(b)$. Im expected to use implicit differentiation to show this property.
The question in its current form isn't correct. If we define $h(x)=\ln(x)+7$, then $h'(x)=\frac{1}{x}$, but 
$$h(ab)=\ln(ab)+7,$$
$$h(a)+h(b)=\ln(a)+7+\ln(b)+7=\ln(ab)+14.$$

EDIT: With the extra information you have given, the solution follows readily from the fundamental theorem of calculus
$$f(x)=f(x_0)+\int_{x_0}^xf'(y)dy,$$
where in this case we set $x_0=1$. I can't however think of any way to do this using only 'implicit differentiation'.
The question in its current form isn't correct. If we define $h(x)=\ln(x)+7$, then $h'(x)=\frac{1}{x}$, but 
$$h(ab)=\ln(ab)+7,$$
$$h(a)+h(b)=\ln(a)+7+\ln(b)+7=\ln(ab)+14.$$

EDIT: With the extra information you have given, the solution follows readily from the fundamental theorem of calculus
$$f(x)=f(x_0)+\int_{x_0}^xf'(y)dy,$$
where in this case we set $x_0=1$. I can't however think of any way to do this using only 'implicit differentiation'.