## anonymous 5 years ago the derivative of x^(-2x) is -2x^(-2x) (log(x)+1) why is there that +1?

1. mathteacher1729

See image. 1) Go to http://www.wolframalpha.com 2) type "derivative x^(-2x) Hope this helps. :)

2. anonymous

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

3. anonymous

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

4. anonymous

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

5. anonymous

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

6. anonymous

7. anonymous

but that is where the -2 comes from

8. anonymous

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

9. anonymous

this is a question :)

10. anonymous

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.

11. anonymous

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

12. anonymous

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

13. anonymous

yep, I see

14. anonymous

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

15. anonymous

yes yes, cool technique

16. anonymous

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

17. anonymous

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

18. anonymous

thanks, I see why this helps

19. anonymous

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