nasty, until you note that
f(x) = (x+1)/(2x+1)
f'(x) = -1/(2x+1)^2
Find f'x given
f(x)= 1/(1+(1/(1+(1/x))))
2 answers
Very cute problem.
The answer is as cute!
Look at the table shown in:
http://prnt.sc/atvi6m
f(x) and f'(x) are given on the third row!
If you are in an exam, your teacher may not appreciate the "cute" answer. So let's work it out:
f(x)=1/(1+(1/(1+(1/x))))
If you work out the fractions, you will find that
f(x)=(x+1)/(2x+1)
Apply the quotient rule we solve readily:
f'(x)=[(2x+1)-2(x+1)]/(2x+1)²
=-1/(2x+1)²
If you have the patience to work out the right-hand side of the third line, you will find
f'(x)=-1/[((2x+1)/(x+1))²*((x+1)/x)²*(x)²]
=-1/(2x+1)²
as above.
The answer is as cute!
Look at the table shown in:
http://prnt.sc/atvi6m
f(x) and f'(x) are given on the third row!
If you are in an exam, your teacher may not appreciate the "cute" answer. So let's work it out:
f(x)=1/(1+(1/(1+(1/x))))
If you work out the fractions, you will find that
f(x)=(x+1)/(2x+1)
Apply the quotient rule we solve readily:
f'(x)=[(2x+1)-2(x+1)]/(2x+1)²
=-1/(2x+1)²
If you have the patience to work out the right-hand side of the third line, you will find
f'(x)=-1/[((2x+1)/(x+1))²*((x+1)/x)²*(x)²]
=-1/(2x+1)²
as above.
1 answer