# The Exponential Function ## Advanced Problem 6 ### Andrew Freeman 9-21-20 Live https://hackmd.io/@AEjIFj77Rxyj8Izesy-cPw/BJe66Y8Sw#/ --- This question has two parts. Both parts ask you to prove the same formula, but they do so in very different ways. --- Algebraically show that the following identity is true $$\frac{(ab)^h−1}{h}=\frac{b^h(a^h−1)+(b^h−1)}{h}$$ Then use this identity along with properties of limits and the definition of m(b) as a limit, to show that for any a,b we have: $$m(ab)=m(a)+m(b)$$ --- $$\frac{(ab)^h−1}{h}=\frac{b^h(a^h−1)+(b^h−1)}{h}$$ $$=\frac{(ab)^h-b^h+b^h-1}{h}$$ $$=\frac{(ab)^h-1}{h}$$ --- It is known that $m(a)=\lim_{h\to 0}\frac{a^h-1}{h}$ and $m(b)=\lim_{h\to 0}\frac{b^h-1}{h}$. These must be added together in order to get $m(ab)$. --- $$m(ab)=\lim_{h\to 0}\frac{a^h-1}{h}+\lim_{h\to 0}\frac{b^h-1}{h}$$ $$m(ab)=\lim_{h\to 0}\frac{b^h(a^h-1)+(b^h-1)}{h}$$ $$m(ab)=\lim_{h\to 0}\frac{(ab)^h-1}{h}$$ --- Consider the formula $(ab)^x=a^xb^x$. By taking derivatives with respect to x on both sides of this formula, applying the appropriate derivative rules, and doing any needed simplifications, prove the following identity for any a,b: $m(ab)=m(a)+m(b)$. --- First find the derivatives of $a^x$, $b^x$, and $(ab)^x$. $$\frac{d}{dx}a^x=m(a) a^x$$ $$\frac{d}{dx}b^x=m(b) b^x$$ $$\frac{d}{dx}(ab)^x=m(ab)(ab)^x$$ --- Now find the derivative for the whole equation and simplify. $$m(ab)(ab)^x=a^x(m(b) b^x)+b^x(m(a) a^x)$$ $$m(ab)(ab)^x=(ab)^x(m(a)+m(b))$$ $$m(ab)=m(a)+m(b)$$ ---