## Advanced Level Presentaion-Section 7.1 ## Anna Smith ### 09/21/2020 --- Question 6: This question has two parts. Both parts ask you to prove the same formula, but they do so in very different ways. --- Part 1: Algebraically show that the following identity is true. $\frac{((ab)^h)-1}{h} =\frac{b^h(a^h-1)+(b^h-1)}{h}$ Multiply both sides by h. $h[\frac{((ab)^h)-1}{h} =\frac{b^h(a^h-1)+(b^h-1)}{h}]h$ This gives you... --- $(ab)^h-1=b^h(a^h-1)+(b^h-1)$ $(ab)^h-1=b^h*a^h-b^h+b^h-1$ Simplyfying this you get... $a^h*b^h-1=a^h*b^h-1$ --- 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: $m(ab)=m(a)+m(b)$ --- Using the limit definition: $m(b)=\lim{h\to0}(\frac{b^h-1}{h})$ --- $m(ab)=\lim{h\to0}(\frac{(ab)^h-1))}{h}=\lim{h\to0}[b^h(\frac{a^h-1}{h})+\frac{b^h-1}{h}]=m(ab)$ $=\lim{h\to0}b^h*\lim{h\to0}(\frac{a^h-1}{h})+ \lim{h\to0}(\frac{b^h-1}{h})$ Limit of a sum=sum of limits. Limit of a product=product of limits. $\lim{h\to0}b^h=b^0=1$ --- This equals $1*m(a)+m(b)=m(a)+m(b)$ or... $m(a)+m(b)=m(a)+m(b)$ --- Consider the formula $(ab)^x=a^x*b^x$. By taking the 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 we use the equation given... $\frac{d}{dx}b^x=b^xm(b)$ by the definition in the book Apply this to a,b... $\frac{d}{dx}(ab)^x=(ab)^xm(ab)$ --- $\frac{d}{dx}(ab)^x=(ab)^xm(ab)=m(ab)a^x*b^x$ Expand the first part $\frac{d}{dx}(ab)^x$ $\frac{d}{dx}a^x*b^x$ --- Use the product rule on the first part: Let $f(x)=a^x$ Let $g(x)=b^x$ $a^x\frac{d}{dx}b^x+\frac{d}{dx}a^x*b^x$ Remember $\frac{d}{dx}b^xm(b)$ apply this to the $a$ and $b$ $[(a^x*b^xm(b)+a^xm(a)*b^x=m(ab)a^x*b^x)]\frac{1}{a^xb^x}$ Final answer: $m(a)+m(b)=m(ab)$