Find the derivative.
f(x)=(1+logᴈx)/x
f'(x)=[1-ln3-logᴈx (ln3)]/(x²ln3)
How do you simplify it further? Thanks.
14 years ago
14 years ago
I would change the log3x to
lnx/ln3
then y = (1 + lnx/ln3)/x
= 1/x + lnx/(ln3x)
dy/dx = -1/x^2 + [(ln3)x((1/x) - x(ln3)]/x^2
= -1/x^2 + [(ln3)x(1/x) - (lnx)(ln3)]/((ln3)^2(x^2))
leaving any further simplification up to you
14 years ago
sh: Your "subscript 3 symbol" showed up as a square on my computer. Hence the confusion.
14 years ago
I understand now, thanks! I changed the log to, 1/xln3 instead.
11 months ago
To simplify the expression [1-ln3-logᴈx(ln3)]/(x²ln3), you can start by combining like terms in the numerator.
First, distribute the negative sign to the two terms within the parentheses:
[1-ln3-(ln3)logᴈx]/(x²ln3)
Then, to combine the two terms with ln3, you can factor out ln3:
[1 - (1 + logᴈx)]/(x²ln3)
Simplifying further, you get:
[1 - 1 - logᴈx]/(x²ln3)
The first two terms cancel out:
[-logᴈx]/(x²ln3)
Finally, you can simplify further by cancelling out the ln3 terms in both the numerator and denominator:
-f'(x) = logᴈx / x²
So, the simplified derivative of f(x) is:
f'(x) = -logᴈx / x²