## calyne 2 years ago Differentiate the function y = 2x*log[10](sqrt(x))

1. calyne

y' = [(2x)/(ln(10)*sqrt(x)] + [2*log[10](sqrt(x))]

2. calyne

correct so far?

3. calyne

or no the derivative of log[10](sqrt(x)) would be (1/[sqrt(x)ln(10)])*(1/[2sqrt(x)])

4. calyne

no it's not

5. calyne

well according to the textbook the answer is pretty different

6. calyne

alright well i've got 1/ln(10) + 2*log[10](sqrt(x))

7. calyne

that's not final though so help me from there

8. calyne

somehow the second term (after the +) needs to equal log[10](x) does that make sense

9. experimentX

isn't it something like this http://www.wolframalpha.com/input/?i=d%2Fdx%28+2x*log+base+10+%28sqrt%28x%29%29%29

10. calyne

sure i guess it's something like that bro jesus you're a great help

11. calyne

but that's still not the textbook answer

12. experimentX

let me test, dy/dx = 2{dx/dx*log[10](sqrt(x))+x*d(log[10](sqrt(x)))/dx} = 2{log[10](sqrt(x))+x*d(ln(sqrt(x)))/dx*1/ln10} = 2{log[10](sqrt(x))+x*1/sqrt(x)*1/2*1/sqrt(x)*1/ln10} = 2{log[10](sqrt(x))+1/2ln10} my answer.

13. calyne

alright here look the textbook answer is 1/ln(10) + log[10](x).

14. experimentX

i got almost same answer ... except 1/ln(10) + 2log[10](x)

15. calyne

i got as far as 1/ln(x) + 2(log[10](x)), the latter of which = 2[log(sqrt(x))/log(10)], so [2*log(sqrt(x))] / log(10) = log(sqrt(x))^2/log(10) ?? which = log(x)/log(10)? which = log[10](x) ???? is that correct? does that make sense? i'm reeeal rusty on the log business

16. calyne

nah bro according to the product rule that other end of the + sign is gonna be the derivative of 2x * the log[10](sqrt(x)). unless that's not the case. but i'm pretty sure it is. don't see any way around it really. and you got the derivative of 2x multiplying the other factor which is supposed to be left as is, so.... you tell me..... if and how i'm wrong.....

17. experimentX

use this property of log to convert it into natural log. $\log_ax = \frac{\ln x}{\ln a}$ I think there's a better example at wikipedia.

18. calyne

i know that, and it still doesn't explain how 2*log[10](sqrt(x)) = log[10](x).

19. calyne

oh unless... that equals log[10]([sqrt(x)]^2) ???? is that it?

20. calyne

oh so it is