Question

the derivative of x^(-2x) is -2x^(-2x) (log(x)+1)

why is there that +1?

Answer 1

See image.

1) Go to http://www.wolframalpha.com
2) type "derivative x^(-2x)

Hope this helps. :)

Answer 2

\[x^{-2x}\] take the log get
\[-2x\ln(x)\]
take the derivative using product rule get
\[-2x\frac{1}{x} -2ln(x)\]

Answer 3

now factor out the 2 get
\[2(\ln(x)-1)\]

Answer 4

of course the x's in the first term cancel. then multiply by the original function to get your answer:
\[x^{-2x}(2(ln(x)-1)=2\times x^{-2x}(ln(x)-1)\]

Answer 5

i made a mistake is see. start with
\[-2-2ln(x)\] gives \[-2(ln(x)+1)\]

Answer 6

sorry about that.

Answer 7

but that is where the -2 comes from

Answer 8

you can do this because ln(x) '=1/x
so if you take the log and at the end you multiply by the function than you get the derrivate

Answer 9

this is a question :)

Answer 10

oh yes. if you want to take the derivative of a function with the variable in the exponent the easiest thing to do is first take the log.

Answer 11

I have never seen this before, but it is a clever idea

Answer 12

you know that the derivative of
\[ln(f(x))\] is
\[\frac{f'(x)}{f(x)}\]

Answer 13

yep, I see

Answer 14

which means you take the log, take the derivative, and then multiply by original function.

Answer 15

yes yes, cool technique

Answer 16

you can also view it this way:
\[x^{2x}=e^{-2x\ln(x)}\]

Answer 17

then take the derivative using the chain rule. the derivative of e to the something is just e to the something times the derivative of something. you will see that the work is identical

Answer 18

thanks, I see why this helps

Answer 19

in any case all the work is taking the derivative of the log of that thing, either way. hope this helps