Implicit differentiation to find property of a function

Question

Im asked to show that if $h(1)=0$ and $h'(x)={1\over{x}}$ then $a,b>0$ show $h(ab)=h(a)+h(b)$. Im expected to use implicit differentiation to show this property.

Answer 1

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'.

Answer 2

Differentiate left hand side $$dh(ab)=\frac{\partial h}{\partial a}b\ da+\frac{\partial h}{\partial b}a\ db=\frac{1}{a\ b}b\ da+\frac{1}{a\ b}a\ db=\frac{1}{a} da+\frac{1}{b} db$$ Differentiate right hand side $$dh(a)+dh(b)=\frac{\partial h}{\partial a} da+\frac{\partial h}{\partial b} db=\frac{1}{a} da+\frac{1}{b} db$$ Both are equal