Here is the answer. I know it is right. The answer you have at the bottom is incorrect.
Can someone please tell me what I'm doing wrong here, and show me how to do it correctly?
I need to differentiate g(x) = LN a-x/a+x
So,
g'(x) = 1/a-x/a+x *d/dx (a-x/a+x)
g'(x) = a+x/a-x *[(a+x)*d/dx(a-x) - (a-x)*d/dx(a+x)/(a+x)^2]
g'(x) = a+x/a-x *[(a+x)*(1-1) - (a-x)(1+1)/(a+x)^2]
g'(x) = a+x/a-x *[(a+x)*(0) - (a-x)(2)/(a+x)^2]
g'(x) = a+x/a-x *[0 - 2(a-x)/(a+x)^2]
g'(x) = a+x/a-x * -2a+2x/a^2+2ax+x^2
g'(x) = a+x*(-2a+2x)/a-x*(a^2+2ax+x^2)
g'(x) = -2a^2+2ax-2ax+2x^2/a^3+2a^2x+ax^2-a^2x-2ax^2-x^3)
g'(x) = -2a^2+2x^2/(a^3+a^2x-ax^2-x^3)
the answer is supposed to be g'(x) = -2a/a^2-x^2, but that's not what I get when I simplify . . .
This is correct...BUT, how you took the derivative is NOT correct.
Again...look at what you have written. It is wrong.
d/dx ln(a-x)= [1/(a-x)]*d/dx[(a-x)]=1/(a-x)]*(-1)=-1/(a-x)
d/dx -ln(a+x)= -[1/(a+x)]*d/dx[(a+x)]=-1/(a+x)]*(+1)=-1/(a+x)
-1/(a-x)-1/(a+x)=(-1(a+x)-1(a-x))/[(a-x)*(a+x)]
=(-a-x-a+x))/[a^2+ax-ax-x^2]=(-a-a)/(a^2-x^2)=-2a/(a^2-x^2)
If you go to my second post where I show the problem aligned in the calculator. Then if you take the derivative of whats on the screen and put it on paper you get my exact answerr. Now maybe this whole scenario revolves around the fact that I didn't understand the way:
g(x)= ln x-a/x+a
There again this is terrible notation because it could me several different things:
g(x)= ln(x-a)/x+a ???
g(x)= ln(x)-a/x+a ???
or g(x)= ln(x-a/x+a) ???
I mean if my professor handed me this in class I would hand it right back to him simply because this in its entirety is "mathematically incorrect". A professor who does this obviously should be teaching grade school or not at all.
Okay, but I don't understand how we're supposed to know that "a" IS a constant term, and x is not. Is there some kind of rule for determining which term is constant when you have two or more varaibles? I thought that constant terms were just integers. For example, in f(x)=x^2-7x+5 the constant term is 5 and when you find f'(x) it's f'(x)=2x-7+0.