In (Root(1-2x))
I personally would differetiate every component using every rule
Firstly the In:
1/Root(1-2x)
and then the Root:
1/2 (1-2x)-1/2
after that the inside (1-2x)
-2
Where I then multiply them all together to get =
- 1/ (1-2x)
This is the correct answer but in the book it never mention such have half-a*s, quick way to find the answer like this. I kidda got lazy and start skipping steps to get this answer. Is this half-a*s? How would you do it, please teach me the correct way.
Thanks.
+118609 Yes you are right although I'm not sure I follow exactly what you are saying.
\(\frac{d}{dx}[lnf(x)]=\frac{f'(x)}{f(x)}\\~\\ if\;\;\\f(x)=(1-2x)^{0.5}\\ then\\ f'(x)=0.5(1-2x)^{-0.5}*-2=-(1-2x)^{-0.5}\\ so\\ \begin{align}\frac{d}{dx}[ln(1-2x)^{0.5}]&=-(1-2x)^{-0.5}\div (1-2x)^{0.5}\\ &=-(1-2x)^{-0.5-0.5}\\ &=\frac{-1}{1-2x} \end{align} \)
