Question

Show the steps for the solutions!

1. Find x. lg 2x=3

2. Find f'(x) when f(x)=1/(1-3x)Show the steps for the solutions!

1. Find x. lg 2x=3

2. Find f'(x) when f(x)=1/(1-3x)@Mathematics

Answer 1

\[\ln 2x = 3\]
Is this the first one?

Answer 2

That is the first problem, yes..

Answer 3

wait, lg means 10 is the base D:

Answer 4

As long as I am concerned, "lg" stands for base 10
\[\large \begin{array}{l}
{\log _{(10)}}2x = 3\\
2x = {10^3}\\
x = \frac{{1000}}{2}\\
x = 500
\end{array}\]

Answer 5

For the second one... you can just use the chain rule.
\[f'(x) = \frac{3}{(1-3x)^2}\]

Answer 6

Apparently they want to use the definition of derivate to solve the second problem. I tried, but got the wrong answer. Can someone show how?

Answer 7

So, they want you to use that stupid limit thing?

Answer 8

I think so :p

Answer 9

but if you were to use the chain rule, how would that be? I'm not taught in english, so I'm not sure what the chain rule is..

Answer 10

Lame...
Remember that it is:
\[f'(x)=\lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h}\]
\[\large\lim_{h\rightarrow0}\frac{\frac{1}{1-3x+h}-\frac{1}{1-3x}}{h}\]
\[\lim_{h\rightarrow0}\frac{1}{-1-h+6 x+3 h x-9 x^2}\]
\[\frac{1}{\lim_{h\rightarrow0}(-1-h+6 x+3 h x-9 x^2)}\]
\[-\frac{1}{(1-3x)^2}\]
Gah... I messed this up somewhere... it's close but not right.

Answer 11

I hate these stupid limits, there are much easier ways of doing them:
The Chain Rule says we can do it this way:
\[f'(x)=-\frac{\frac{d}{dx}(3-3x)}{(1-3x)^2}=\frac{3}{(1-3x)^2}\]

Answer 12

that should be a one in the numerator for the derivative... it doesnt matter though, because any constant goes to 0.

Answer 13

Now I'm just confused D:

Answer 14

Oh, I see where I messed up

Answer 15

\[\large\lim_{h\rightarrow0}\frac{\frac{1}{1-3(x+h)}-\frac{1}{1-3x}}{h}\]
\[\large\lim_{h\rightarrow0}\frac{\frac{1}{1-3x-3h}-\frac{1}{1-3x}}{h}\]
\[\large\lim_{h\rightarrow0}\frac{3}{1-3 h-6 x+9 h x+9 x^2}\]
\[\large\frac{3}{\lim_{h\rightarrow0} (1-3 h-6 x+9 h x+9 x^2)}\]
\[\frac{3}{1-6x+8x^2}\]
\[\frac{3}{(1-3x)^2}\]