Find f'x given
f(x)= 1/(1+(1/(1+(1/x))))
nasty, until you note that
f(x) = (x+1)/(2x+1)
f'(x) = -1/(2x+1)^2
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.
To find the derivative of f(x), we can start by simplifying the function. Let's rewrite f(x) as follows:
f(x) = 1 / (1 + (1 / (1 + (1 / x))))
Now, let's simplify this expression further. To do this, we'll start from the innermost fraction and simplify the expression one step at a time:
1 + (1 / x)
= (x / x) + (1 / x) [Multiplying the first fraction by (x / x)]
= (x + 1) / x
Next, let's substitute this simplified expression back into the original function:
f(x) = 1 / (1 + (1 / ((x + 1) / x)))
Now, let's simplify the expression inside the parentheses:
1 / ((x + 1) / x)
= x / (x + 1) [Inverting the fraction by flipping the numerator and denominator]
Substituting this simplified expression back into the original function:
f(x) = 1 / (1 + (x / (x + 1)))
To find the derivative f'(x), we can use the quotient rule. The quotient rule states that if we have a function of the form f(x) = g(x) / h(x), where g(x) and h(x) are functions, then the derivative f'(x) is given by:
f'(x) = (g'(x) * h(x) - h'(x) * g(x)) / (h(x) ^ 2)
Applying the quotient rule to our function f(x), we have:
f'(x) = ((1 + (x / (x + 1)))' * (1 + (x / (x + 1)))) - (1 * (1 + (x / (x + 1)))') / ((1 + (x / (x + 1))))^2
Now, let's find the derivatives of the individual terms:
((1 + (x / (x + 1)))'
= (1)' + ((x / (x + 1)))' [Sum rule]
= 0 + ((x * (x + 1) - x) / ((x + 1) ^ 2)) [Applying the quotient rule]
= ((x^2 + x - x) / ((x + 1) ^ 2))
= (x^2 / ((x + 1) ^ 2))
(1 + (x / (x + 1)))'
= (1)' + ((x / (x + 1)))' [Sum rule]
= 0 + ((x * (x + 1) - x) / ((x + 1) ^ 2)) [Applying the quotient rule]
= ((x^2 + x - x) / ((x + 1) ^ 2))
= (x^2 / ((x + 1) ^ 2))
Substituting these derivatives back into the equation for f'(x), we have:
f'(x) = (x^2 / ((x + 1) ^ 2) * (1 + (x / (x + 1)))) - (1 * (x^2 / ((x + 1) ^ 2))) / ((1 + (x / (x + 1))))^2
Simplifying this expression further, we have:
f'(x) = (x^2 / ((x + 1) ^ 2) * (1 + (x / (x + 1)))) - (x^2 / ((x + 1) ^ 2)) / ((1 + (x / (x + 1))))^2
And finally, we can simplify this expression further if necessary.