Find the derivative of:
ln(2x/x+1)
the answer is 1/x(x+1) but i can't seem to solve it that far.
ln [2x/(x+1)] = ln 2 + ln (x) - ln(x+1)
Add the derivatives of each term and you get
1/x - 1/(x+1) = [(x+1) - x]/[x(x+1)] = 1/[x(x+1)]
To find the derivative of ln(2x/(x+1)), we can use the properties of logarithms and the quotient rule of differentiation.
Step 1: Rewrite the expression using the properties of logarithms.
ln(2x/(x+1)) = ln(2x) - ln(x+1)
Step 2: Differentiate each term separately.
The derivative of ln(2x) can be found using the chain rule, where the derivative of ln(u) is (1/u) * du/dx.
So, the derivative of ln(2x) is (1/(2x)) * (d(2x)/dx) = 1/x.
The derivative of ln(x+1) can also be found using the chain rule. The derivative of ln(u) is (1/u) * du/dx.
So, the derivative of ln(x+1) is (1/(x+1)) * (d(x+1)/dx) = 1/(x+1).
Step 3: Combine the derivatives.
Using the properties of logarithms, we subtract the derivatives:
(1/x) - (1/(x+1)) = (x+1 - x)/(x(x+1)) = 1/(x(x+1))
Therefore, the derivative of ln(2x/(x+1)) is 1/(x(x+1)).