Morgage payment acording to natural logrithmics

[rboatwr0]

The term t (in years) of a $200,000 home mortgage at 7.5% interest can be approximated by the function below, where x is the monthly payment in dollars.

t = 13.375 ln

x

x − 1250

, x[FONT=verdana, helvetica, sans-serif] > 1250[/FONT]
[FONT=verdana, helvetica, sans-serif]

The part of the problem I am stuck on says
[/FONT](d) Find the instantaneous rates of change of [FONT=verdana, helvetica, sans-serif]t with respect to [/FONT][FONT=verdana, helvetica, sans-serif]x when [/FONT][FONT=verdana, helvetica, sans-serif]x = [/FONT]$1394.09[FONT=verdana, helvetica, sans-serif] and [/FONT][FONT=verdana, helvetica, sans-serif]x = [/FONT]$1618.17[FONT=verdana, helvetica, sans-serif]. (Round your answers to four decimal places.)[/FONT]

[FONT=verdana, helvetica, sans-serif]Next I found t'[/FONT]

[FONT=verdana, helvetica, sans-serif]13.375 (lnx - ln(x-1250)[/FONT]
[FONT=verdana, helvetica, sans-serif]13.375(1/x - 1/(x-1250))[/FONT]

[FONT=verdana, helvetica, sans-serif]plugged in 1394.09 and got the answer -.0832 which was wrong.... is my derivative correct? If so what did I do wrong?

Thanks in advance![/FONT]

 

[stapel]

The term t (in years) of a $200,000 home mortgage at 7.5% interest can be approximated by the function below, where x is the monthly payment in dollars.

t = 13.375 ln

x

x − 1250

,x > 1250

Did you mean "t = 13.375 ln[x / (x - 1250)], for x > 1250"? (Math-formatting advice available here.)

. . . . .\(\displaystyle t\, =\, 13.375\ln\left(\dfrac{x}{x\, -\, 1250}\right)\)

I found t'

13.375 (lnx - ln(x-1250)
13.375(1/x - 1/(x-1250))

Does the above perhaps represent "t=", followed by your derivative of t?

plugged in 1394.09 and got the answer -.0832 which was wrong.... is my derivative correct? If so what did I do wrong?

The derivative returns negative values; perhaps the original function contains an error...?

 

[rboatwr0]

Did you mean "t = 13.375 ln[x / (x - 1250)]

Yes that's what I meant. It looked fine when I hit send! I'm not sure why it changed.

Also by t' I was stating my d/dt=13.375(1/x - 1/(x-1250))

I got this by using the property of ln(a/b)=ln(a)-ln(b) and d/dx ln(u)= (1/u)*u'

 

[stapel]

The derivative of the posted function returns negative values. Are you sure the original function is posted correctly?

 

[rboatwr0]

The derivative of the posted function returns negative values. Are you sure the original function is posted correctly?

Yes. Im sure. And negative would make sense because if you pay more on a loan, the time you have to pay it will shrink