Quick quadrature question.

[Jack_O]

I have an equation in the form y=10^x, i have an error in x, dx. I am unsure how to find the associated error in y. dy=10^dx gives to small an error and dy=x-dx doesn't seem logical as a smaller error in x gives a bigger error in y.

(The actual equation is d=10^((m(v)-M(v)+5)/5), i have already have errors for m(v) and M(v) and have combined them by using sqrt(dM(v)^2+dm(v)^2), the equation is used for calculating cosmoligical distances).

 

[tiny-tim]

I have an equation in the form y=10^x, i have an error in x, dx. I am unsure how to find the associated error in y.

Hi Jack_O!

(try using the X² tag just above the Reply box )

Hint: just differentiate … if y = 10^x, then dy = (what)dx ?

 

[Jack_O]

Hi tiny-tim, doesn't bode well that i had to look up the differential of 10^x

Anyhoo i know get my error in y as 10^xln(10)dx, which gives me a much more reasonable answer, thanks for your help

 

[tiny-tim]

Hi tiny-tim, doesn't bode well that i had to look up the differential of 10^x

Hi Jack_O!

Quick trick: 10^x = (e^ln10)^x = e^xln10